I found
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx=I\tag1.$$
Mathematica failed to find $I$, so I am not sure if there is closed form for it. I am just giving it a try here.
First idea came to my mind is to use the Fourier series of $-\ln(\cos x)=\ln(2)+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$ and we have
$$I=8\ln(2)\underbrace{\int_0^{\pi/2}x^2\cot x\ dx}_{\frac32\ln(2)\zeta(2)-\frac78\zeta(3)}+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x^2 \cot x\cos(2nx)\ dx.$$
I got stuck here. Any help would be much appreciated.
Proof of $(1)$
from here we have
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
replace $x$ by $\sqrt{x}$ we get
$$\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$
multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=2\int_0^1\frac{\arcsin^2(\sqrt{x})\ln(1-x)}{x}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx$$
Best Answer
$$S=-8 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{1}{3} \log ^4(2)+\frac{4}{3} \pi ^2 \log ^2(2)$$
Proof $1$. This. Proof $2$. This. Proof $3$. This. Bonus: $$\small \int_0^{\frac{\pi }{2}} x^3 \cot (x) \log (\cos (x)) \, dx=\frac{3}{2} \pi \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{16} \pi \zeta (3) \log (2)-\frac{\pi ^5}{120}+\frac{1}{16} \pi \log ^4(2)-\frac{1}{8} \pi ^3 \log ^2(2)$$