Inspired by me

The sum is $\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2} \; = \; 1 + \sum_{n=1}^\infty \frac{H_{n+1}^{(3)}}{(n+1)^2} \; = \; 1 + \sum_{n=1}^\infty \frac{H_n^{(3)}}{(n+1)^2} + \sum_{n=1}^\infty \frac{1}{(n+1)^5} \; = \; \zeta(5) + \sigma_h(3,2)$ using the standard notation for the Euler sum $\sigma_h(m,n) \; = \; \sum_{k=1}^\infty \left(\sum_{j=1}^k \frac{1}{j^m}\right)\frac{1}{(k+1)^n} \; =\; \sum_{k=1}^\infty \frac{H_k^{(m)}}{(k+1)^n} \;.$ Now standard results tell us that $\sigma_h(3,2) \; = \; \tfrac92\zeta(5) - 2\zeta(2)\zeta(3)$ and hence the sum is $\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2} \; = \; \tfrac{11}{2}\zeta(5) - 2\zeta(2)\zeta(3) \; = \; \tfrac{11}{2}\zeta(5) - \tfrac13\pi^2\zeta(3)$ making the answer $11+5+2+2+3+3 = \boxed{26}$ .