Explicit value for series We consider the sum
$$
\sum_{n_{1},\, n_{2}\ \in\ \mathbb{Z}}\,\,\,
\frac{16}{(2 n_1 - 2 \nu n_2 + 1)^2 (2 n_1 + 2 \nu - 2 \nu n_2 - 1)^2}
$$
with $\nu = (-1)^{1/3}$ and $\mathrm{i}$ is the imaginary unit.

*

*I am wondering if there is a closed form expression for that double series.

*Notice, that it definitely converges, as the denominator essentially behaves like
$\left\Vert\,{n}\,\right\Vert^{4}$.

 A: Let $L=\Bbb{Z}+e^{2i\pi /3}\Bbb{Z}$, your sum is $f(1/2)$ with
$$f(z)=\sum_{l\in L} \frac1{(z+l)^2(z+l+e^{2i\pi/3}-1)^2}$$
Which is meromorphic $L$-periodic with a pole of order $2$, with $a=e^{2i\pi/3}-1$
we get $$f(z)=\frac{2}{a^2}\wp_L(z)+C$$
Next, substracting from $\wp_L'(z)^2-4\wp_L(z)^3$ a sum over its poles to get a bounded entire function, we have $$\wp_L'(z)^2=4\wp_L(z)^3-\frac43 \pi^4 E_4(e^{2i\pi /3}) \wp_L(z)-\frac8{27} \pi^6 E_6(e^{2i\pi /3})$$ where $\frac43 \pi^4 E_4(e^{2i\pi /3}),\frac8{27} \pi^6 E_6(e^{2i\pi/3})$ are given in term of the Laurent coefficients of $\wp_L(z)-z^{-2}$. Because $L=e^{2i\pi /3}L$ we have $E_4(e^{2i\pi /3})=0$. And $\wp_L'(z)$ is odd $L$-periodic thus $\wp_L'(1/2)=\wp_L'(-1/2)=0$.
Whence
$$\wp_L(1/2) =  e^{2i\pi n/3} (\frac14 \frac8{27} \pi^6 E_6(e^{2i\pi /3}))^{1/3}$$ Finally, this is where we need the whole theory of complex tori/elliptic curves/modular forms, $$E_6(e^{2i\pi /3})=(-1728 \Delta(e^{2i\pi/3}))^{1/2}=\pm i 3\sqrt{3} 2^3 (2\pi)^6 \eta(e^{2i\pi/3})^{12}$$ $$= \pm i 3\sqrt{3} 2^3 (2\pi)^6  (e^{-\frac{\pi i}{24}}\frac{\sqrt[8]{3}  \Gamma \left(\frac{1}{3}\right)^{3/2}}{2 \pi })^{12}$$

Computation of $C$:
$\sum_m \sum_n$ means $\lim_{M\to \infty} \sum_{m=-M}^M \lim_{N\to \infty} \sum_{n=-N}^N$.
From $\sum_n \frac1{(z+\nu^t n)^2}=\nu^{-2t}\frac{\pi^2}{\sin^2(\pi \nu^{-t} z)}$ and the absolute convergence of $\sum_m \frac{\pi^2}{\sin^2(\pi (\nu^{-t} z+m \nu))}$
then $$g(z)=\sum_{t=0}^2 \sum_m \sum_n \frac1{(z+\nu^t (n+m \nu))^2}+\frac1{(z+\nu^t(n+m \nu)+a)^2}$$ $$=2\sum_{t=0}^2  \sum_m \sum_n \frac1{(z+\nu^t (n+m \nu))^2}$$ $$
=2\sum_m \sum_n \sum_{t=0}^2 \frac1{(z+\nu^t (n+m \nu))^2}-\frac{1_{(n,m)\ne (0,0)}}{(\nu^t (n+m \nu))^2}=6 \wp_L(z)$$
From $\sum_n \frac1{z+n}-\frac1{z+n+b}=\frac{2i\pi}{e^{2i\pi z}-1}-\frac{2i\pi}{e^{2i\pi (z+b)}-1}$ then
$$h(z)=\sum_{t=0}^2 \sum_m \sum_n \frac1{z+\nu^t (n+m\nu)}-\frac1{z+\nu^t (n+m\nu)+a}$$ $$=\sum_{t=0}^2 \nu^{-t}\sum_m \frac{2i\pi}{e^{2i\pi (\nu^{-t} z+m\nu )}-1}-\frac{2i\pi}{e^{2i\pi (\nu^{-t} (z+a)+m\nu )}-1}=2i\pi (\nu^0+\nu^{-1}-2\nu^{-2})=-6i\pi \nu$$
and hence $$3 f(z)=
3\sum_{l\in L} \frac1{a^2 (z+l)^2} - \frac2{a^3 (z+l)} + \frac1{a^2 (a + z+l)^2} + \frac2{a^3 (a + z+l)}$$ $$=\sum_{t=0}^2\sum_{l\in L} \frac1{a^2 (z+\nu^t l)^2} - \frac2{a^3 (z+\nu^t  l)} + \frac1{a^2 (a + z+\nu^t  l)^2} + \frac2{a^3 (a + z+\nu^t  l)}$$ $$=
\frac1{a^2} g(z)-\frac2{a^3} h(z)=\frac6{a^2} \wp_L(z)+\frac2{a^3} 6i\pi \nu$$
