2D lattice sum with numerator

The $n=4$ sum is accessible as follows. Expand $(j+k)^4$; by symmetry the odd terms $4j^3k$ and $4jk^3$ cancel out so we need only sum $(j^4 + 6j^2k^2 + k^4) / (j^2+k^2)^4$. You say you already know the sum of $1 / (j^2+k^2)^2$, which is $(j^4 + 2j^2k^2 + k^4) / (j^2+k^2)^4$ [see also the "P.S." paragraph below]; so we only need one more sum $\sum^\prime (j^4 + c j^2 k^2 + k^4) / (j^2+k^2)^4$. Well, we know the sum of $1 / (j+ik)^4$ from the theory of the Weierstrass $\wp$-function; if I did this right it comes to $\Gamma(1/4)^8 / (960 \pi^2) = 3.1512120+$. But $$ \frac1{(j+ik)^4} = \frac{(j-ik)^4}{j^2+k^2)^4} = \frac{(j^4 - 4i j^3 k - 6 j^2 k^2 + 4i j k^3 + k^4}{(j^2+k^2)^4}, $$ and the odd terms again cancel out, so this gives us the sum of $(j^4 - 6j^2k^2 + k^4) / (j^2+k^2)^4$ — and we're done because $$ 2 (j^4 + 6j^2k^2 + k^4) = 3 (j^2+k^2)^2 - (j^4 - 6j^2k^2 + k^4). $$

P.S. In general the sum of $1/(j^2+k^2)^n$ is $4 \zeta(n) L(n,\chi_4)$ [because it's 4 times the value at $n$ of the zeta function for ${\bf Q}(i)$]. Thus when $n=2$ that's $2\pi^2 {\bf G} / 3$ where ${\bf G}$ is Catalan's constant $$ L(2,\chi_4) = 1 - \frac1{3^2} + \frac1{5^2} - \frac1{7^2} + - \cdots = .915965594\ldots $$ so $2\pi^2 {\bf G} / 3 = 6.026812\ldots$; no further simplification is known or expected, though there are various ways known to compute the numerical value of ${\bf G}$ efficiently to any desired accuracy.