$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
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\newcommand{\imp}{\Longrightarrow}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\sum _{n = 1}^{\infty}{\pars{n!}^{2} \over n^{3}\pars{2n}!}} & =
\sum _{n = 1}^{\infty}{1 \over n^{3}}\,
{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} =
\sum _{n = 1}^{\infty}{1 \over n^{2}}\,\ \overbrace{%
{\Gamma\pars{n + 1}\Gamma\pars{n} \over \Gamma\pars{2n + 1}}}
^{\ds{\mathrm{B}\pars{n + 1,n}}}
\\[4mm] & =
\sum _{n = 1}^{\infty}\
{1 \over n^{2}}\ \overbrace{\int_{0}^{1}x^{n}\pars{1 - x}^{n - 1}\,\dd x}
^{\ds{\mathrm{B}\pars{n + 1,n}}}\ =\
\int_{0}^{1}{1 \over x}\sum_{n = 1}^{\infty}
{\bracks{x\pars{1 - x}}^{\, n} \over n^{2}}\,\dd x
\\[4mm] & =
\int_{0}^{1}{\,\mathrm{Li}_{2}\pars{x\bracks{1 - x}} \over x}\,\dd x =
-\int_{0}^{1}\ln\pars{x}\,\mathrm{Li}_{2}'\pars{x\bracks{1 - x}}\pars{1 - 2x}
\,\dd x
\\[4mm] & =
-\int_{0}^{1}\ln\pars{x}\,
\braces{-\,{\ln\pars{1 - x\bracks{1 - x}} \over x\pars{1 - x}}}
\pars{1 - 2x}\,\dd x
\\[4mm] & =
\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x -
\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over 1 - x}\,\dd x
\\[4mm] & =
\underbrace{\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x}
_{\ds{\,\mathcal{I}_{1}}}\ -\
\underbrace{\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x}
_{\ds{\,\mathcal{I}_{2}}}
\\[4mm] & = \,\mathcal{I}_{1} - \,\mathcal{I}_{2}\tag{1}
\end{align}
1. $\ds{\large\,\mathcal{I}_{1} =\ ?}$.
\begin{align}
\,\mathcal{I}_{1} & =
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad%
\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad}
\\ \mbox{}\\ \hline
\end{array} =
\int_{0}^{1}\ln\pars{x}\ln\pars{1 + x^{3} \over 1 + x}\,{\dd x \over x}
\\[4mm] & =
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{3}} \over x}\,\dd x -
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x
\\ & =
{1 \over 9}\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x -
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x =
-\,{8 \over 9}
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x
\\[4mm] & =
{8 \over 9}
\int_{0}^{-1}\bracks{-\,{\ln\pars{1 - x} \over x}}\,\ln\pars{-x}\,\dd x =
{8 \over 9}
\int_{0}^{-1}\,\mathrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x =
-\,{8 \over 9}
\int_{0}^{-1}{\,\mathrm{Li}_{2}\pars{x} \over x}\,\dd x
\\[4mm] & =
-\,{8 \over 9}\int_{0}^{-1}\,\mathrm{Li}_{3}'\pars{x}\,\dd x =
-\,{8 \over 9}\,\mathrm{Li}_{3}\pars{-1} =
-\,{8 \over 9}\pars{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}} -
\sum_{n = 1}^{\infty}{1 \over n^{3}} +
\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}}
\\[4mm] & =
{2 \over 3}\sum_{n = 1}^{\infty}{1 \over n^{3}}
\end{align}
\begin{equation}
\,\mathcal{I}_{1} =
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad%
\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad}
\\ \mbox{}\\ \hline
\end{array} =
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad{2 \over 3}\,\zeta\pars{3}\quad}
\\ \mbox{}\\ \hline
\end{array}\tag{2}
\end{equation}
2. $\ds{\large\,\mathcal{I}_{2} =\ ?}$.
\begin{align}
\,\mathcal{I}_{2} & =
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad%
\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad}
\\ \mbox{}\\ \hline
\end{array} =
-\int_{0}^{1}\,\mathrm{Li}_{2}'\pars{x}\ln\pars{x^{2} - x + 1}\,\dd x
\\[4mm] & =
\int_{0}^{1}\,\mathrm{Li}_{2}\pars{x}\,{2x - 1 \over x^{2} - x + 1}\,\dd x
\end{align}
The roots of $\ds{x^{2} - x + 1}$ are given by
$\ds{r = {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$ and $\ds{\ol{r}}$ such that
\begin{align}
{2x - 1 \over x^{2} - x + 1} & =
2\pars{{x - 1/2 \over x - r} - {x - 1/2 \over x - \ol{r}}}\,{1 \over r - \ol{r}}
=
2\,{1 \over 2\ic\,\Im\pars{r}}\,2\ic\,\Im\pars{x - 1/2 \over x - r}
\\[5mm] & =
{2 \over \root{3}}\,\Im\pars{2r - 1 \over x - r} = 2\,\Re\pars{1 \over x - r}
\end{align}
\begin{align}
\,\mathcal{I}_{2} & =
2\,\Re\bracks{\int_{0}^{1}{\,\mathrm{Li}_{2}\pars{x} \over x - r}\,\dd x} =
-2\,\Re\bracks{\int_{0}^{1}
{\,\mathrm{Li}_{2}\pars{x} \over 1 - x/r}\,{\dd x \over r}} =
-2\,\Re\bracks{\int_{0}^{\ol{r}}
{\,\mathrm{Li}_{2}\pars{rx} \over 1 - x}\,\dd x}
\\[5mm] & =
2\,\Re\bracks{\left.\vphantom{\LARGE A}\ln\pars{1 - x}\,\mathrm{Li}_{2}\pars{rx}\,
\right\vert_{\ 0}^{\ \ol{r}} -
\int_{0}^{\ol{r}}\ln\pars{1 - x}\,\mathrm{Li}_{2}'\pars{rx}r\,\dd x}
\\[5mm] & =
2\,\Re\int_{0}^{\ol{r}}\ln\pars{1 - x}\,{\ln\pars{1 - rx} \over x}\,\dd x
\\[5mm] & =
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x +
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - rx} \over x}\,\dd x -
\Re\int_{0}^{\ol{r}}\ln^{2}\pars{1 - x \over 1 - rx}\,{\dd x \over x}
\\[5mm] & =
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x +
\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x -
\Re\int_{0}^{\ol{r}}\ln^{2}\pars{1 - x \over 1 - rx}\,{\dd x \over x}
\end{align}
In the last integral, we'll make the
sub$\ds{\ldots {1 - rx \over 1 - x} \mapsto x}$ such that the whole integration is reduced to:
\begin{align}
\,\mathcal{I}_{2} & =
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x +
\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x - \bracks{%
\Re\int_{1}^{0}{\ln^{2}\pars{x} \over -1 + x}\,\dd x -
\Re\int_{1}^{0}{\ln^{2}\pars{x} \over -r + x}\,\dd x}
\\[5mm] & =
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x +
\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{rx} \over 1 - x}\,\dd x
\end{align}
The remaining integrals can be evaluated by 'standard means'. For instance
$$
\left\lbrace\begin{array}{rcl}
\ds{\int{\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} &
\ds{\ln^{2}\pars{1 - x}\ln\pars{x} + 2\ln\pars{1 - x}
\,\mathrm{Li}_{2}\pars{1 - x} - 2\,\mathrm{Li}_{3}\pars{1 - x}}
\\[3mm]
\ds{\int{\ln^{2}\pars{rx} \over 1 - x}\,\dd x} & \ds{=} &
\ds{-\ln\pars{1 - x}\ln^{2}\pars{rx} - 2\ln\pars{rx}
\,\mathrm{Li}_{2}\pars{x} + 2\,\mathrm{Li}_{3}\pars{x}}
\end{array}\right.
$$
$$
\left\lbrace\begin{array}{rcl}
\ds{\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} &
\ds{-\,{2\pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r} -
2\Re\,\mathrm{Li}_{3}\pars{r} + 2\
\overbrace{\mathrm{Li}_{3}\pars{1}}^{\ds{\zeta\pars{3}}}}
\\[3mm]
\ds{\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{rx} \over 1 - x}\,\dd x} & \ds{=} &
\ds{2\,\Re\,\mathrm{Li}_{3}\pars{\ol{r}}}
\end{array}\right.
$$
With these results:
\begin{equation}
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad\,\mathcal{I}_{2} =
2\zeta\pars{3} - {2 \pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r}\,,\qquad
r = \half + {\root{3} \over 2}\,\ic = \expo{\pi\ic/3}\quad}
\\ \mbox{}\\ \hline
\end{array}\tag{3}
\end{equation}
With $\ds{\pars{1}, \pars{2}\ \mbox{and}\ \pars{3}}$, we'll arrive to the $\ds{\underline{final\ result}}$:
$$
\begin{array}{|rcl|}\hline\mbox{}\\
\ds{\quad\color{#f00}{\sum _{n = 1}^{\infty}{\pars{n!}^{2} \over n^{3}\pars{2n}!}}}
& \ds{=} &
\ds{\color{#f00}{{2 \pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r} -
{4 \over 3 }\,\zeta\pars{3}}}\ds{\,\, \approx\,\, 0.5229\quad}
\\[3mm]
\ds{r} & \ds{\equiv} & \ds{\expo{\pi\ic/3} = \half + {\root{3} \over 2}\,\ic}
\\ \mbox{}\\ \hline
\end{array}
$$
In order to see the relation with the Trigamma Function $\ds{\Psi\, '}$, we can exploit the 'symmetries' of $\ds{\expo{\pi\ic/3}}$. Namely,
\begin{align}
\Im\,\mathrm{Li}_{2}\pars{r} & =
\Im\,\mathrm{Li}_{2}\pars{\expo{\pi\ic/3}} =
\Im\sum_{n = 1}^{\infty}{\expo{n\pi\ic/3} \over n^{2}} =
\Im\sum_{n = 0}^{\infty}\bracks{%
{\expo{\pars{3n + 1}\pi\ic/3} \over \pars{3n + 1}^{2}} +
{\expo{\pars{3n + 2}\pi\ic/3} \over \pars{3n + 2}^{2}} +
{\expo{\pars{3n + 3}\pi\ic/3} \over \pars{3n + 3}^{2}}}
\\[5mm] & =
{\root{3} \over 2}\bracks{%
\sum_{n = 0}{\pars{-1}^{n} \over \pars{3n + 1}^{2}} +
{\pars{-1}^{n} \over \pars{3n + 2}^{2}}}
\end{align}
However,
\begin{align}
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{3n + z}^{2}} & =
\sum_{n = 0}^{\infty}{1 \over \pars{6n + z}^{2}} -
\sum_{n = 0}^{\infty}{1 \over \pars{6n + z + 2}^{2}}
\\[5mm] & =
{1 \over 36}\sum_{n = 0}^{\infty}{1 \over \pars{n + z/6}^{2}} -
{1 \over 36}\sum_{n = 0}^{\infty}{1 \over \pars{n + z/6 + 1/3}^{2}}
\\[5mm] & =
{1 \over 36}\,\Psi\, '\pars{z \over 6} -
{1 \over 36}\,\Psi\, '\pars{{z \over 6} + {1 \over 3}}
\end{align}
Then,
\begin{align}
\Im\,\mathrm{Li}_{2}\pars{r} & =
{\root{3} \over 72}\bracks{%
\Psi\, '\pars{1 \over 6} -\
\underbrace{\Psi\,'\pars{\half}}_{\ds{\pi^{2} \over 2}}\ +\
\Psi\, '\pars{1 \over 3} - \Psi \, '\pars{2 \over 3}}
\end{align}
