I have been working in this series $$\sum _{m=0}^{\infty } \sum _{k=0}^{\infty } \sum _{j=0}^{\infty } \frac{(-1)^{j+k+m} \left((j+1)^2+(k+1)^2+(m+1)^2\right)}{\left((j+1)^2+(k+1)^2+(m+1)^2\right)^{3/2}}=0.132476$$ I used the Nsum as usual in Mathematica and it gives the result above, using different values of working precision I get similar answer, but a few different decimals.
Maple it is not capable of giving any value. I used another method and it gives similar values as above but this represents the madelung constant and it must be $\frac{-1,73.}{2}$ for the extended sum to -$\infty$. I do not know where is the error? Mathematica is wrong or the Levin Accelerator method is wrong too.
Your sum is $$ S=\sum_{j,k,m=0}^{\infty}\frac{(-1)^{j+k+m}}{\sqrt{(j+1)^2+(k+1)^2+(m+1)^2}}=-\sum_{j,k,m=1}^{\infty}\frac{(-1)^{j+k+m}}{\sqrt{j^2+k^2+m^2}}, $$ while the Madelung constant is $$ M={\sum_{j,k,m=-\infty}^{\infty}}^\prime\frac{(-1)^{j+k+m}}{\sqrt{j^2+k^2+m^2}}, $$ with the $^\prime$ indicating that the term with $j=k=m=0$ is to be skipped. Note that your sum accounts for one of the $8$ octants in the Madelung sum (the one with all positive coordinates), but skips all the terms with any coordinate equal to zero, and has an extra minus sign. Filling in the missing terms, you can say that $$ M=-8S+12\sum_{j,k=1}^{\infty}\frac{(-1)^{j+k}}{\sqrt{j^2+k^2}}+6\sum_{j=1}^{\infty}\frac{(-1)^j}{j}. $$ The final sum is $-\log 2\approx -0.693147$, and the middle sum is (numerically) something like $0.2892 \pm 0.0001$; putting these together with OP's $S\approx 0.132476$ gives $M\approx -1.748 \pm 0.001$, which is close enough ... it agrees with the reported value of $M$, which is $-1.747565$ according to Wikipedia.