Is the closed form for $\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}$ known in the literature?
Yes, a closed form in the literature is known. For $m \geqslant 2$ it is: $$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$ Here $\eta (s) = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^s} = (1 - 2^{1 - s}) \zeta (s)$ is the Dirichlet eta function and $\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s}$ is the Riemann zeta function.
References:
R. Sitaramachandrarao, "A formula of S. Ramanujan," Journal of Number Theory, 25, 1-19 (1987). See Theorem 3.5 on page 9.
Philippe Flajolet and Bruno Salvy, "Euler sums and contour integral representations," Experimental Mathematics, 7(1), 15-35 (1998). See Theorem 7.1 (i) on page 32.
The value of the series may be extracted from Theorem $1$ of the preprint A simple strategy of calculating two alternating harmonic series generalizations. More precisely, we have
Let $m\ge2$ be a positive integer. The following equalities hold: \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x \end{equation*} \begin{equation*} \small =\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr), \end{equation*} where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$ and $\zeta$ denotes the Riemann zeta function.
One may also check https://math.stackexchange.com/q/3236584.
A note: the series mentioned by omegadot (from a paper by R. Sitaramachandrarao, "A formula of S. Ramanujan," Journal of Number Theory, 25, 1-19 (1987). See Theorem 3.5 on page 9) seems to be strongly related to the series above if we look at their integral representations, which will be mentioned in the next version of the paper.