using the fact that $\displaystyle\sum_{n=1}^\infty z^nH_n^{(3)}=\frac{\operatorname{Li}_3(z)}{1-z}\ $ divide both sides by $z$ then integrate from $z=0$ to $x$, we get
$$\sum_{n=1}^\infty \frac{x^nH_n^{(3)}}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\tag{1}$$
replace $x$ with $-x$ in(1), then divide both sides by $x$ and integrate from $0$ to $1$, we get
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}&=\operatorname{Li}_5(-1)-\underbrace{\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}\ dx}_{IBP}-\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\
&=\operatorname{Li}_5(-1)+\operatorname{Li}_2(-1)\operatorname{Li}_3(-1)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\
&=\frac38\zeta(2)\zeta(3)-\frac{15}{16}\zeta(5)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\
\end{align}
I proved here
\begin{align}
\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx=\frac34\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5)
\end{align}
which follows that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$
BONUS:
By setting $x=-1$ in (1) we have
$$\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$