My question is at the end of the solution.
We have
$$\int_0^1 x^{n-1}\ln(1-x)=-\frac{H_n}{n}$$
Differentiate both sides with respect to $n$
$$\int_0^1 x^{n-1}\ln x\ln(1-x)=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$
Next multiply both sides by $\frac{4n}{n{2n\choose n}}$ then $\sum_{n=1}^\infty$ we get
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(4x)^n}{n{2n\choose n}}\right)dx$$
$$=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{2\sqrt{x}\arcsin\sqrt{x}}{\sqrt{1-x}}\right)dx$$
$$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta=16I$$
For the integral, let $\theta\to \frac{\pi}{2}-\theta$ to have
$$I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta-\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta$$
$$\Longrightarrow 2I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta$$
$$=\frac{\pi}{2}\left(\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{48}\right)$$
$$\Longrightarrow I=\frac34\ln^2(2)\zeta(2)-\frac{15}{32}\zeta(4)$$
Therefore
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{2}\zeta(4)$$
Since
$$\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\zeta(2)\left(\frac{\pi^2}{2}\right)=\frac{15}{2}\zeta(4)$$
we have the nice relation
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)$$
Finally substitute
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
we obtain
$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\approx 6.2957$$
But Mathematica gives
$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.04326$$
Can you spot any mistake or my solution is good?
Thank you.
@User 628759 gave the solution. It is quite surprising to see how sensitive it the result to this parameter.
On my side, I computed exactly $$\sum_{n=1}^{10000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.23740$$ $$\sum_{n=1}^{20000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.25448$$
To be slow, it is !