Is there a closed form for $\sum_{n=1}^\infty\frac{(-4)^nH_{n-1}^3}{{2n\choose n}n^2}\ ?$ How to evaluate 

$$S=\sum_{n=1}^\infty\frac{(-4)^nH_{n-1}^3}{{2n\choose n}n^2}\ ?$$
  where $H_n$ is the harmonic number. 

This problem was posted on a Facebook group with no answer so I am not sure if there is a closed form but seems an interesting problem to try.
I know that from here we have
$$\small{\sum_{n=1}^\infty H_{n-1}^3x^{n-1}=
\frac{\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)}{1-x}}$$ 
and from here we have 
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
My question is can we exploit the two identities above to get $S$ or is there a different way to get $S$?
Thank you,
I am tagging "integration" as most series can be converted to integrals. 
 A: For sake of completing my comment above, I hereby present my unsuccessful attempt:
Use the formula I commented, which is easily provable from the Kronecker delta function $\delta_{mn}=\frac{1}{2\pi}\int_0^{2\pi}e^{\pm i(m-n)t}dt$, and use the two formulae the OP provided, we get:
\begin{align*}
S&=\frac{1}{2\pi}\int_0^{2\pi}\left(\sum_{m=1}^\infty\frac{\left(4e^{-it}\right)^m}{m^2\binom{2m}{m}}\right)\left(\sum_{n=1}^\infty\left(-e^{it}\right)^nH_{n-1}^3\right)dt\\
&=-\frac{1}{\pi}\int_0^{2\pi}\arcsin^2\left(e^{-\frac{it}{2}}\right)e^{it}\frac{F\left(-e^{it}\right)}{1+e^{it}}dt
\end{align*}
where 
$$F(x)=\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)$$
Then, by a variable change $x=e^{\frac{it}{2}}$ and noting that the integrand is an even function with respect to $x$, the integral reduces to
\begin{align*}
S&=\frac{4}{\pi}\Im\int_0^1x\arcsin^2\left(\frac{1}{x}\right)\frac{F\left(-x^2\right)}{1+x^2}dx\\
&=-2\int_0^1\log\left(\frac{1+\sqrt{1-x}}{\sqrt{x}}\right)\frac{F\left(-x\right)}{1+x}dx
\end{align*}
as confirmed by Mathematica numerically. 
