I managed to derive the following generalization
$$\sum_{n=1}^\infty\frac{H_{\frac np}}{n^q}=(-1)^qp \sum_{n=1}^\infty\frac{H_{pn}}{(pn)^q}-\sum_{j=1}^{q-2}(-p)^{-j}\zeta(q-j)\zeta(j+1)\tag1$$
and wondering if its known in the literature or not.
One of the nice applications is
$$\sum_{n=1}^\infty\frac{H_{\frac n2}}{n^{2q}}=\left[1+2^{-2q-1}(2q+1)\right]\zeta(2q+1)-\frac12\sum_{j=1}^{2q-2}\left[1-(-2)^{1-j}\right]\zeta(j+1)\zeta(2q-j)$$ $$+\sum_{j=1}^{q-1}\left[1-2^{1-2j}\right]\zeta(2j)\zeta(2q-2j+1)\tag2$$
which follows from setting $p=2$ and replacing $q$ by $2q$ then substituting the two generalizations: $\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2q}}$ and $\sum_{n=1}^\infty \frac{H_n}{n^q}$.
Another application is $$\sum_{n=1}^\infty\frac{H_n}{n^{q}}=-\frac12\sum_{i=1}^{q-2}(-1)^i\zeta(q-i)\zeta(i+1),\quad q=3,5,7,...$$
which follows from setting $p=1.$
Thanks
Here is the proof of $(1)$.
$$\sum_{n=1}^\infty\frac{H_{\frac np}}{n^q}=\frac1p\sum_{n=1}^\infty\frac{1}{n^{q-1}}\left(\frac{H_{\frac np}}{\frac np}\right)$$
use that $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ $$=\frac1p\sum_{n=1}^\infty\frac{1}{n^{q-1}}\left(-\int_0^1 x^{\frac np-1}\ln(1-x)dx\right)$$
$$=-\frac1p\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{x^{\frac np}}{n^{q-1}}dx$$
$$=-\frac1p\int_0^1\frac{\ln(1-x)}{x}\text{Li}_{q-1}(x^{\frac 1p})dx$$
set $x^{\frac 1p}=y$ $$=-\int_0^1 \frac{\ln(1-y^p)\text{Li}_{q-1}(y)}{y}dy$$
expand $\ln(1-y^p)$ in series $$=\sum_{n=1}^\infty\frac{1}{n}\int_0^1 y^{pn-1}\text{Li}_{q-1}(y)dy$$
integrate by parts repeatedly $$=\sum_{n=1}^\infty\frac 1n\left((-1)^q\frac{H_{pn}}{(pn)^{q-1}}-\sum_{j=1}^{q-2}(-1)^j\frac{\zeta(q-j)}{(pn)^j}\right)$$
$$=(-1)^qp\sum_{n=1}^\infty\frac{H_{pn}}{(pn)^q}-\sum_{j=1}^{q-2}(-p)^j\zeta(q-j)\left(\sum_{n=1}^\infty\frac{1}{n^{j+1}}\right)$$
$$=(-1)^qp\sum_{n=1}^\infty\frac{H_{pn}}{(pn)^q}-\sum_{j=1}^{q-2}(-p)^j\zeta(q-j)\zeta(j+1).$$
By the way, using $(2)$, we can also find the two useful generalizations:
$$\sum_{n=1}^\infty\frac{(-1)^nH_{\frac n2}}{n^{2q}}=\left[2^{-2q-1}(2q+3)-1\right]\zeta(2q+1)$$ $$-\frac12\sum_{j=1}^{2q-2}\left[2^{1-2q}-1+(-2)^{1-j}\right]\zeta(2q-j)\zeta(j+1)-\sum_{j=1}^{q-1}\left[1-2^{1-2j}\right]\zeta(2j)\zeta(2q-2j+1)$$
and
$$\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=\left[q-2^{-2q-1}(2q+1)\right]\zeta(2q+1)+\left[2^{1-2q}-2\right]\ln(2)\zeta(2q)$$ $$-\frac12\sum_{j=1}^{2q-2}(-2)^{1-j}\zeta(j+1)\zeta(2q-j)-\sum_{j=1}^{q-1}\left[1-2^{1-2j}\right]\zeta(2j)\zeta(2q-2j+1)$$
