How to show that: \begin{aligned} & f(z) = \frac{1}{(2z+1)^{3}\sin(\pi z)} \\ \sum_{n=-\infty}^{\infty} \operatorname{Res}[f(z), n] = 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)^{3}} \end{aligned}?
When we consider (as $\lim_{z\mapsto n}$ for $\operatorname{Res}$ doesn't work.
On
$f(z)$ is a meromorphic function with a triple pole at $z=-\frac{1}{2}$ and a simple pole at any $z=n\in\mathbb{Z}$.
The residue at $-\frac{1}{2}$ equals $-\frac{\pi^2}{16}$ and by De l'Hopital theorem
$$ \operatorname*{Res}_{z=n}f(z) =\lim_{z\to n}\frac{1}{(2z+1)^3\pi\cos(\pi z)}=\frac{(-1)^n}{\pi(2n+1)^3}.$$
This implies that
$$ \sum_{n\in\mathbb{Z}}\operatorname*{Res}_{z=n}f(z) = 2\sum_{m\geq 0}\frac{(-1)^m}{\pi(2m+1)^3} = -\operatorname*{Res}_{z=-1/2}f(z) = \frac{\pi^2}{16} $$
thus
$$ \sum_{m\geq 0}\frac{(-1)^m}{(2m+1)^3}=\frac{\pi^3}{32}.$$
I think there is a problem with your answer. There should be a multiple of $\pi$ in the denominator of the result you give, i.e. $$\sum_{n=-\infty}^{\infty}Res[f(z),n]=2\cdot \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3\cdot \pi}$$ Because:
For every $n$, $f$ has a simple pole. Hence, just use the following formula ($f$ and $g$ in the following formula are general functions):
$$Res[f/g,n]=f(n)/g^{\prime}(n)$$
So, we have
$$Res[\frac{1}{(2z+1)^3\cdot sin(\pi z)},n]=\frac{1}{3\cdot (2n+1)^2\cdot 2\cdot sin(\pi n)+(2n+1)^3\cdot cos(\pi n) \cdot \pi}=\frac{(-1)^n}{(2n+1)^3\cdot \pi}$$ Now, observe that $n$ and $-n-1$ give the same result. Because: $$(2n+1)^3\cdot \pi=-(-2n-1)^3\cdot \pi$$ since $x^3$ is an odd function. Moreover, if $n$ is odd, $-n-1$ is even and vice versa. Therefore, indeed, $n$ and $-n-1$ give the same result.
Hence, we have the desired result.