Is the result for $3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}$ known in the literature? I was able to get the following result

$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}=11\zeta(3)\zeta(6)+\frac52\zeta(4)\zeta(5)-\frac{13}{6}\zeta^3(3)-2\zeta(2)\zeta(7)-5\zeta(9)$$
  where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

based on a nice identity and some manageable Euler sums.  Is this result known in the literature?  Can we evaluate the terms separately?

In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al. 
The results are:
$$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta (6) + \frac{173}{72} \zeta (9) + \frac{1}{4} \zeta (4) \zeta (5) - 3 \zeta (2) \zeta (7) - \frac{2}{3} \zeta^3 (3) \quad \text{(See Eq. 3.18)}$$
and
$$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^5} = \frac{679}{24} \zeta (9) - 11 \zeta (2) \zeta (7) - \frac{1}{2} \zeta (3) \zeta (6) - \frac{29}{4} \zeta (4) \zeta (5) - \frac{1}{6} \zeta^3 (3).$$
