Here it is my approach, with some definitions first: $\mathscr{H}_n$ is $\sum_{k=0}^{n}\frac{1}{2k+1}$; $D^{1/2}$ is the semi-derivative operator, i.e. a linear operator which acts on power series (centered at $0$ or $1$) by mapping $x^{\alpha}$ into $x^{\alpha-1/2}\frac{\Gamma(\alpha+1)}{\Gamma(\alpha+1/2)}$; $D^{1/2}_{\perp}$ is the adjoint operator of $D^{1/2}$ with respect to the standard inner product of $L^2(0,1)$, which is denoted through $\langle\cdot,\cdot\rangle$; $K$ is Catalan's constant; $K(x)$ and $E(x)$ are the complete elliptic integrals of the first and second kind, with the parameter being the elliptic modulus. $\tau$ is the involutive and self-adjoint operator bringing $g(x)$ into $g(1-x)$; $\text{IBP},\text{SBP}$ stand for integration/summation by parts; $\text{FL}$ stands for Fourier-Legendre expansion(s).
$\newcommand{Li}{\operatorname{Li}}$
By semi-integration by parts $\zeta(2)$ equals
$$ \int_{0}^{1}\frac{\arcsin(\sqrt{x})\arcsin(\sqrt{1-x})}{\sqrt{x(1-x)}}\,dx=\frac{\pi}{2}\int_{0}^{1}D^{1/2}\log(1-x)\cdot D^{1/2}_{\perp}\log(x)\,dx. $$
where the LHS can be written both in terms of power series and Fourier-Legendre expansions, leading to:
$$ \begin{eqnarray*}\frac{\pi^3}{24}&=&\sum_{m,n\geq 0}\frac{\left[\frac{1}{4^m}\binom{2m}{m}\right]\left[\frac{1}{4^n}\binom{2n}{n}\right]}{(2n+1)(2m+1)(m+n+1)\binom{m+n}{n}}\\ &=& \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\pi(-1)^n+\frac{2}{(2n+1)}-4(-1)^n\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right]^2\end{eqnarray*}$$
or, equivalently:
$$\begin{eqnarray*} \frac{\pi^3}{24}&=&\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\frac{2}{(2n+1)}+4(-1)^n\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\\&=&\frac{\pi^3}{8}+16\sum_{n\geq 0}\frac{1}{(2n+1)^2}\sum_{k>n}\frac{(-1)^k}{2k+1}+16\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\end{eqnarray*}$$
$$-\frac{\pi^3}{192}=-\sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{(2n+3)}+\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2$$
$$\begin{eqnarray*}&& \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\\ &=& \frac{5\pi^3}{192}-\frac{\pi K}{4}+K-\frac{\pi^2}{16}+\frac{\pi\log(2)}{4}-\int_{0}^{1}\sum_{m\geq 0}\frac{x^{2m}(-1)^m}{(2m+1)}\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\,dx\end{eqnarray*} $$
$$\begin{eqnarray*} \sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2n+1)^2 (2m+1) (2m+2n+1)}&\stackrel{\text{sym}}{=}&\sum_{m,n\geq 0}\frac{(-1)^{m+n}(m+n+1)}{(2m+1)^2(2n+1)^2(2m+2n+1)}\\
&=&\frac{K^2}{2}+\frac{1}{2}\int_{0}^{1}\left(\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\right)^2\,dx
\end{eqnarray*}$$
via
$$ \frac{\arcsin(\sqrt{x})}{\sqrt{x}}=\sum_{n\geq 0}P_n(2x-1)\left[\pi(-1)^n+\frac{2}{(2n+1)}-4(-1)^n\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right].$$
By summation by arts and Fourier-Legendre expansions we have the following relations:
$$ \sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{(2n+1)}\stackrel{\text{SBP}}{\longleftrightarrow}\sum_{n\geq 0}\frac{\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}}{(2n+1)^2}
\stackrel{\text{FL}}{\longleftrightarrow}\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx=\left\langle D^{-1/2}\frac{\text{arctanh}{\sqrt{x}}}{x\sqrt{\pi}},K(1-x)\right\rangle $$
where the RHS equals
$$\begin{eqnarray*} \left\langle \frac{\text{arctanh}{\sqrt{x}}}{x\sqrt{\pi}},(\tau D^{-1/2})K(x)\right\rangle &=& \int_{0}^{1}\frac{\text{arctanh}{\sqrt{x}}\arcsin(\sqrt{1-x})}{x}\,dx\\ &=& 2\int_{0}^{\pi/2}\theta\tan(\theta)\text{arctanh}{\cos\theta}\,d\theta\end{eqnarray*} $$
and
$$ \int_{0}^{\pi/2}\theta \tan(\theta)\left(\cos\theta\right)^{2k+1}\,d\theta \stackrel{\text{IBP}}{=}\frac{4^k}{(2k+1)^2 \binom{2k}{k}} $$
allows us to state
$$\begin{eqnarray*}\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx&=&2\sum_{k\geq 0}\frac{4^k}{(2k+1)^3 \binom{2k}{k}}=-2\int_{0}^{1}\frac{\arcsin(x)\log(x)}{x\sqrt{1-x^2}}\,dx\\&=&2\int_{0}^{\pi/2}\frac{-\theta\log(\sin\theta)}{\sin(\theta)}\,d\theta.\end{eqnarray*}$$
We may notice that
$$ -\log(\sin \theta) = \log(2)+\sum_{n\geq 1}\frac{\cos(2n\theta)}{n},$$
$$ \int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\cos(2n\theta)\,d\theta = 2K-2\sum_{k=0}^{n-1}\frac{(-1)^k}{(2k+1)^2}, $$
hence the integral $\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx$ boils down to the Euler sum
$$ \sum_{n\geq 1}\frac{1}{n}\sum_{k\geq n}\frac{(-1)^k}{(2k+1)^2}\stackrel{\text{SBP}}{=}\sum_{n\geq 1}\frac{(-1)^n H_n}{(2n+1)^2}=\int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx $$
getting rid of some alternating Stirling numbers. Additionally the last integral is known (at least) since De Doelder and Flajolet:
$$ \int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx = -\frac{\pi^3}{64}-K\log(2)-\frac{\pi}{16}\log^2(2)+2\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right).$$
Now we just have to re-combine all the pieces to find a closed form for $\int_{0}^{1}\operatorname{Ti}_2(x)^2\,\frac{dx}{x^2}$ and the monstrosity $\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left(\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)}\right)^2$.
$$ \sum_{k\geq 0}\frac{4^k}{(2k+1)^3 \binom{2k}{k}} =\phantom{}_4
F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)
= -\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2)+4\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$
$$ \sum_{n\geq
0}\frac{1}{(2n+1)^2}\left[\pi+\frac{2(-1)^n}{(2n+1)}-4\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right]=-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2)+4\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$
$$ \sum_{n\geq
0}\frac{1}{(2n+1)^2}\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}=\frac{7\pi^3}{128}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$
$$ \sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{2n+3} =
\frac{3\pi^3}{128}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)
$$
$$ \sum_{n\geq
0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2=\frac{7\pi^3}{384}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)
$$
$$ \sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2n+1)^2 (2m+1)
(2m+2n+1)}=\frac{\pi^3}{128}-\frac{\pi
K}{4}+K-\frac{\pi^2}{16}+\frac{\pi\log(2)}{4}-\frac{\pi\log^2(2)}{32}+\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)
$$
$$ \int_{0}^{1}\left(\sum_{n\geq
0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\right)^2\,dx
=\frac{\pi^3}{64}-\frac{\pi K}{2}+2K-K^2-\frac{\pi^2}{8}+\frac{\pi\log(2)}{2}-\frac{\pi\log^2(2)}{16}+2\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right).$$
