Computing $\sum _{ k=1 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { k\left( k+n \right) }^{ 2 } } } } $

$$ \begin{align} \sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2} &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{kn^2}\tag{1}\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac1{kn^2}\tag{2}\\ &=\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{3} \end{align} $$ Explanation:
$(1)$: substitute $n\mapsto n-k$
$(2)$: change order of summation
$(3)$: use the definition of $H_n$

We could finish by applying this answer with $q=2$, but let's do a bit more work. $$ \begin{align} \sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2} &=\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k(k+n)^2}\tag{4}\\ &=\zeta(3)+\frac12\left(\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k(k+n)^2}+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n(k+n)^2}\right)\tag{5}\\ &=\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{nk(n+k)}\tag{6}\\ &=\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n^2}\left(\frac1k-\frac1{n+k}\right)\tag{7}\\ &=\zeta(3)+\frac12\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{8} \end{align} $$ Explanation:
$(4)$: pull out the $n=0$ terms as $\zeta(3)$
$(5)$: swap $n\leftrightarrow k$ by symmetry
$(6)$: add the summands
$(7)$: partial fractions
$(8)$: $H_n=\sum\limits_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)$

Two times $(8)$ minus $(3)$ yields $$ \sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2}=2\zeta(3)\tag{9} $$

The inner sum is the polygamma function of order $1$:

$$\psi^{(1)}(k)=\sum_{n=0}^{\infty}\frac{1}{(k+n)^2}\tag{1}$$

So the double sum can be written as

$$\sum_{k=1}^{\infty}\frac{\psi^{(1)}(k)}{k}=2\zeta(3)\approx2.404\tag{2}$$

where $\zeta(s)$ is the Riemann Zeta function. I obtained the latter result using WolframAlpha.