Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$, we get from this post that apparently,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{\,k+1}(z)$$
for $-1\leq z\leq 1$, and with Nielsen generalized polylogarithm $S_{n,p}(z)$ and polylogarithm $\rm{Li}_n(z)$. Hence for small $k$,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2\, 2^n}= S_{1,2}\big(\tfrac12\big)+\rm{Li}_3\big(\tfrac12\big)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3\, 2^n}= S_{2,2}\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^4\, 2^n}= S_{3,2}\big(\tfrac12\big)+\rm{Li}_5\big(\tfrac12\big)$$
and so on. Explicitly, given $a=\ln 2$,
$$S_{1,2}\big(\tfrac12\big) +\tfrac1{6}a^3-\tfrac18 \zeta(3)=0 $$
$$S_{2,2}\big(\tfrac12\big) +\tfrac1{168}a^4+\tfrac17a^2\,\rm{Li}_2\big(\tfrac12\big)+\tfrac17a\,\rm{Li}_3\big(\tfrac12\big)-\tfrac18\zeta(4) = 0$$
which are discussed in this and this post. And by yours truly,
$$S_{3,2}\big(\tfrac12\big) -A+B = 0$$
$$A = \tfrac{41}{840}a^5+\tfrac5{21}a^3\,\rm{Li}_2\big(\tfrac12\big)+\tfrac47a^2\,\rm{Li}_3\big(\tfrac12\big)+a\,\rm{Li}_4\big(\tfrac12\big) + \rm{Li}_5\big(\tfrac12\big) $$
$$B=\tfrac12\zeta(2)\zeta(3)+\tfrac18a\,\zeta(4)-\tfrac1{32}\zeta(5)$$
Q: What, however, is the explicit evaluation in ordinary polylogs of the next steps, namely $S_{4,2}\big(\tfrac12\big)$ and $S_{5,2}\big(\tfrac12\big)$?
P.S. Try as I might, they resist being evaluated and there are indications these higher order integrals may not be expressible by ordinary polylogs.