Evaluating the challenging sum $\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$. I managed to evaluate the sum, my approach can be found $\underline{\operatorname{below as an answer}}$, I'd truly appreciate if any of you could share new methods to evaluate this series, thank you.
The following are the short proofs of the other series I encountered during my approach.
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x}}-1\tag{$\ast$}$$
$$\sum _{k=1}^{\infty }\frac{x^k}{k\:4^k}\binom{2k}{k}=\:-2\ln \left(1+\sqrt{1-x}\right)+2\ln \left(2\right)$$

$$\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}=\:2\ln \left(2\right)$$

On $\left(\ast\right)$ perform the following manipulations.
$$-\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(x\right)\:dx=-\int _0^1\frac{\ln \left(x\right)\left(1-\sqrt{1-x}\right)}{x\sqrt{1-x}}\:dx$$
$$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=-2\int _0^1\frac{\ln \left(1-x^2\right)}{1+x}\:dx$$
$$=-2\int _0^1\frac{\ln \left(1-x\right)}{1+x}\:dx-2\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$
$$=2\operatorname{Li}_2\left(\frac{1}{2}\right)-\ln ^2\left(2\right)$$

$$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=\zeta \left(2\right)-2\ln ^2\left(2\right)$$

 A: First part.

Evaluating $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx$$

$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{1}{2}\int _0^1\ln ^2\left(t\right)\left(\int _0^1\frac{1}{\left(1-tx\right)\sqrt{1-x^2}}\:dx\right)dt$$
$$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\left(t\right)}{\sqrt{1-t^2}}\:dt+\frac{1}{2}\int _0^1\frac{\ln ^2\left(t\right)\arcsin \left(t\right)}{\sqrt{1-t^2}}\:dt$$
$$=\frac{\pi }{4}\int _0^{\frac{\pi }{2}}\ln ^2\left(\sin \left(x\right)\right)\:dx+\frac{1}{2}\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\:dx$$
Using the Beta function for the first integral and using this result for the $2$nd one proved there we conclude that:
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{41}{64}\zeta \left(4\right)+\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)+\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{48}\ln ^4\left(2\right)$$

Second part.

Finding $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$

By employing $1+\sum _{k=1}^{\infty }\frac{x^{2k}}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x^2}}$ on the previous integral we have:
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x}\:dx+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\operatorname{Li}_3\left(x\right)\:dx$$
$$=\zeta \left(4\right)+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\left(\frac{1}{2k}\zeta \left(3\right)-\frac{1}{2k}\int _0^1x^{2k-1}\operatorname{Li}_2\left(x\right)\:dx\right)$$
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\zeta \left(4\right)+\frac{1}{2}\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}-\frac{1}{4}\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}$$
$$+\frac{1}{8}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=8\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx-3\zeta \left(4\right)-8\ln \left(2\right)\zeta \left(3\right)-4\ln ^2\left(2\right)\zeta \left(2\right)$$
Using the result found on the first part we finalize.
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=\frac{17}{8}\zeta \left(4\right)+4\operatorname{Li}_4\left(\frac{1}{2}\right)-8\ln \left(2\right)\zeta \left(3\right)+4\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^4\left(2\right)$$
