General solution
I have found a systematic path to a general solution.
It turns out the the generalized problem is easier to solve and that, as a by-product, it reveals clearly the structure of the problem.
So for natural $m$ consider the sum
$$s_{m} = \sum_{n=1}^\infty\frac{(-1)^n\,H_{n/m}}n\tag{1}$$
Doing the $k$-sum after replacing $H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{m}$ we can write
$$\sum _{k=1}^{\infty } \frac{(-1)^k \left(1-x^{k/m}\right)}{k (1-x)}=\int_0^1 \frac{-\log (2)+\log \left(1+x^{1/m}\right)}{1-x}\,dx
\\\overset{x\to t^m}=-\int_0^1 m t^{m-1}\left(\frac{\log(2)-\log(1+t)}{1-t^m} \right)\,dt
\\
=\log(2)\log(1-t^m) + m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\tag{2}$$
Here in the last line we have switches to indeinite integrals to avoid spurious divergencies.
Integration by parts of the last integral with
$$u = \int m \frac{t^{m-1}}{1-t^m}\,dt = - \log(1-t^m), v = \log(1+t)\tag{3}$$
gives
$$m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\\
= - \log(1-t^m)\log(1+t) + \int \frac{\log(1-t^m)}{1+t}\,dt\tag{4}$$
Collecting what we have up to now in
$$f_{m}(t) = - \log(1-t^m)\log(\frac{1+t}{2}) + \int \frac{\log(1-t^m)}{1+t}\,dt \tag{5}$$
so that
$$s_m = f_{m}(t\to 1)-f_{m}(t\to0)= \int_0^1 \frac{\log(1-t^m)}{1+t}\,dt\tag{6}$$
because the product of the logs vanishes at both ends. For what follows it will be convenient technically to proceed with the indefinite version $f_m(t)$.
The integral can be calculated. Writing
$$1-t^m = \prod_{k=1}^{m} (\alpha_{m,k} - t)$$
where the roots are
$$\alpha_{m,k} = \exp(2 i \pi \frac{k}{m}), k=1..m\tag{7}$$
In what follows we will drop the index $m$ in $\alpha$ for simplicity.
Now the log can be written as
$$\log(1-t^m) = \log(\prod_{k=1}^{m}(\alpha_{m,k}-t))= \sum_{k=1}^{m}\log(\alpha_{m,k}-t)\tag{8}$$
and the integral boils down to terms of the form
$$\int \frac{\log (\alpha -t)}{t+1} \, dt =\operatorname{Li}_2\left(\frac{\alpha -t}{\alpha +1}\right)+\log \left(\frac{t+1}{\alpha +1}\right) \log (\alpha -t)\tag{9}$$
Now observing that we can write
$$- \log(1-t^m)\log\left(\frac{1+t}{2}\right)= -\log\left(\frac{1+t}{2}\right)\sum_{k=1}^{m} \log(\alpha_k - t)\tag{10}$$
the indefinite integral $(5)$ can be written as
$$f_{m}(t) = \sum_{k=1}^m \left(\operatorname{Li}_2\left(\frac{\alpha_{k} -t}{\alpha_{k} +1}\right)
\\
+\log \left(\frac{t+1}{\alpha_{k} +1}\right) \log (\alpha_{k} -t)-\log\left(\frac{1+t}{2}\right) \log(\alpha_k - t)\right)\tag{11}$$
To make this expression definite (and finite) we follow $(6)$ and take the value at $t=1$ and subtract its value at $t=0$
$$\phi_{m}(t) = f_{m}(t)- f_{m}(0)$$
This results in a sum over
$$A_{k} = \operatorname{Li}_2\left(\frac{\alpha_{k}-1}{\alpha_{k}+1}\right)-\operatorname{Li}_2\left(\frac{\alpha_{k}}{\alpha_{k}+1}\right)-\log\left(1-\frac{1}{\alpha_{k}}\right)\log\left(\frac{1+\alpha_{k}}{2}\right)\tag{12}$$
There's a still little technical difficulty to circumvent: it turns out that inserting the value of the index $k$ as an integer leads to a divergent behaviour of some $A_{k}$.
Hence we understand $A_{k}$ as the limit from below $A_{k}=\lim_{\epsilon\to +0} \,A_{k-\epsilon}$
Now we are ready to write down the closed expression for the sum $(1)$
$$s_{m} = \sum_{k=1}^{m} A_{k}\tag{13}$$
The general structure of the solution is clear: the sum is composed of quadratic $\log$s and linear $\operatorname{Li}_2$s. No higher polylogs appear.
Remark 1: The main remaining task would be a possible simplification of the polylog function $\operatorname{Li}_2(z)$ of a composite and complex arguments.
Remark 2: It seems possible to do a similar analysis for any rational $m=\frac{p}{q}$.
Remark 3: as we have encountered only antiderivatives before the final limits, we can apply the same reasoning to find the generating function
$$g_{m}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k} H_{\frac{k}{m}}$$
Evaluation of specific cases
Now we turn to the harvest. But I ask for some patience.
Original post
With the help of Mathematica I find a long expression which is numerically correct but needs plenty of simplification in order to become "culturally acceptable".
Doing the $k$-sum after replacing $H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{5}$ gives the integral
$$i= \int_0^1 \frac{\log (2)-\log \left(\sqrt[5]{x}+1\right)}{x-1} \, dx$$
Letting $x\to t^5$ the integrand of $dt$ becomes
$$\frac{5 t^4 \log \left(\frac{2}{t+1}\right)}{t^5-1}=\frac{5 t^4 \log (2)}{t^5-1}-\frac{5 t^4 \log (t+1)}{t^5-1}$$
To avoid divergences we continue with the indefinite integral.
The first integral is
$$i_1= \int \frac{5 t^4 \log (2)}{t^5-1} = \log (2) \log \left(t^5-1\right)$$
and the second one is done by Mathematica returning a lengthy expression.
The we form the sum $i=i_1 + i_2$ and calculate the limits $t\to0$ and $t\to1$ to find this horrendeous expression
$$i = -\text{Li}_2\left(\frac{\sqrt[5]{-1}}{-1+\sqrt[5]{-1}}\right)+\text{Li}_2\left(\frac{1+\sqrt[5]{-1}}{-1+\sqrt[5]{-1}}\right)-\text{Li}_2\left(\frac{(-1)^{2/5}}{1+(-1)^{2/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{2/5}}{1+(-1)^{2/5}}\right)-\text{Li}_2\left(\frac{(-1)^{3/5}}{-1+(-1)^{3/5}}\right)+\text{Li}_2\left(\frac{1+(-1)^{3/5}}{-1+(-1)^{3/5}}\right)-\text{Li}_2\left(\frac{(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\frac{43 \pi ^2}{60}+\frac{\log ^2(2)}{2}+\log (5) \log (2)-\log \left(\sqrt[5]{-1}-1\right) \log \left(1+\sqrt[5]{-1}\right)-\log \left((-1)^{2/5}-1\right) \log \left(1+(-1)^{2/5}\right)-\log \left((-1)^{3/5}-1\right) \log \left(1+(-1)^{3/5}\right)-\log \left((-1)^{4/5}-1\right) \log \left(1+(-1)^{4/5}\right)+\frac{1}{5} i \pi \left(\log \left(\sqrt[5]{-1}-1\right)+5 \log \left(1+\sqrt[5]{-1}\right)+2 \log \left(1+(-1)^{2/5}\right)+3 \log \left((-1)^{3/5}-1\right)+5 \log \left(1+(-1)^{3/5}\right)+4 \log \left(1+(-1)^{4/5}\right)\right)$$
The numeric values of which is $N(i) = -0.152666...$