A much easier approach:
By Cauchy product we have
$$-\ln(1-x)\operatorname{Li}_2(x)=\sum_{n=1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{3}{n^3}\right)x^n$$
replace $x$ with $-x$ then multiply both sides by $-\frac{\ln x}{x}$ and integrate between $0$ and $1$ plus use the fact that $\int_0^1-x^{n-1}\ln x\ dx=\frac1{n^2}$ we get
$$2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-3\operatorname{Li}_5(-1)=\int_0^1\frac{\ln(1+x)\operatorname{Li}_2(-x)\ln x}{x}dx$$
$$\overset{IBP}{=}\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}dx=\frac{5}{16}\zeta(2)\zeta(3)+\frac{7}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$
where the last result follows from this solution, check Eq$(3)$.
rearrange to get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}=\frac{5}{16}\zeta(2)\zeta(3)-\frac{45}{16}\zeta(5)+\frac{7}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$
substitute $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ and $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$, we get
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)$$
Bonus:
Again, by Cauchy product we have
$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$
set $x=-1$ and substitute the result of $\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^4}$ we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$
Or it can be found here.