Here is my slight variation on how I go about things. As you will see, the ideas used and results found are almost the same as yours.
I will make use of result (1) you kindly provide in your answer, namely
$$\sum_{n = 1}^\infty \frac{H^{(3)}_n x^n}{n} = \operatorname{Li}_4 (x) - \ln (1 - x) \operatorname{Li}_3 (x) - \frac{1}{2} \operatorname{Li}^2_2 (x)\tag1$$
Since
$$\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n},$$
one can express the sum as
$$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} = \sum_{n = 1}^\infty \frac{H^{(3)}_n}{n} \cdot \frac{H_n}{n} = -\int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{H^{(3)}_n x^n}{n} \, dx\tag2$$
Substituting (1) into (2) leads to
\begin{align}
\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} &= - \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_4 (x)}{x} \, dx + \int_0^1 \frac{\ln^2 (1 - x) \operatorname{Li}_3 (x)}{x} \, dx\\
& \qquad + \frac{1}{2} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx\\
&= -I_1 + I_2 + \frac{1}{2} I_3.
\end{align}
The first integral $I_1$
\begin{align}
I_1 &= \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_4 (x)}{x} \, dx}_{IBP \,\, 3 \,\, \text{times}}\\
&= -\sum_{n = 1}^\infty \left [\frac{\zeta (4)}{n^2} - \frac{\zeta (3)}{n^3} + \frac{\zeta (2)}{n^4} + \frac{1}{n^4} \int_0^1 x^{n - 1} \ln (1 - x) \, dx \right ]\\
&= -\zeta (4) \sum_{n = 1}^\infty \frac{1}{n^2} + \zeta (3) \sum_{n = 1}^\infty \frac{1}{n^3} - \zeta (2) \sum_{n = 1}^\infty \frac{1}{n^4} + \sum_{n = 1}^\infty \frac{H_n}{n^5}\\
&= -\zeta (4) \zeta (2) + \zeta^2 (3) - \zeta (2) \zeta (4) + \frac{7}{4} \zeta (6) - \frac{1}{2} \zeta^2 (3)\\
&= \frac{1}{2} \zeta^2 (3) - \frac{7}{4} \zeta (6),
\end{align}
where the results
$$\sum_{n = 1}^\infty \frac{H_n}{n^5} = \frac{7}{4} \zeta (6) - \frac{1}{2} \zeta^2 (3) \quad \text{and} \quad \zeta (2) \zeta (4) = \frac{7}{4} \zeta (6),$$
were used.
The second integral $I_2$
In this question here I showed that
$$I_2 = 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \zeta^2 (3) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \zeta (2) \zeta (4) + 2 \sum_{n = 1}^\infty \frac{H^2_n}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}.$$
All four of the sums appearing in the above expression for $I_2$ are known. The first, second, and fourth sums are standard Euler sums while a proof of the third sum can be found here. The results are:
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n^2} &= 2 \zeta (2)\\
\sum_{n = 1}^\infty \frac{H_n}{n^3} &= \frac{5}{4} \zeta (4)\\
\sum_{n = 1}^\infty \frac{H_n}{n^5} &= -\frac{1}{2} \zeta^2 (3) + \frac{7}{4} \zeta (6)\\
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \frac{97}{24} \zeta (6) - 2 \zeta^2 (3),
\end{align}
Thus
$$I_2 = -\zeta^2 (3) + \frac{89}{24} \zeta (6).$$
The third integral $I_3$
\begin{align}
I_3 &= \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx}_{IBP}\\
&= -\operatorname{Li}^3_2(1) - 2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx\\
\Rightarrow I_3 &= -\frac{1}{3} \zeta^3 (2) = -\frac{35}{24} \zeta (6).
\end{align}
The main sum
Combining the results found for the above three integrals, for the sum one has
$$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} = \frac{227}{48} \zeta (6) - \frac{3}{2} \zeta^2 (3),$$
as required.