Maybe seeing that the paranthesis can be factorized as $3/2((m_1+1)^2+(m_2+1)^2+(m_1+m_2)^2)-2$ might help?

Unless it's the sphere or other "simple" manifold, like the comments above say, there are no explicit expressions.

Excluding the examples from the previous answer does not mean the inequality is true for the remaining cases. It is possible that you need to exclude other possibilites of the form $a_1 n_1+a_2n_2+a_3n_3=0$ with $a_i$ integers.

@Neal: If I'm not mistaken it gives the same relation as the second one in the question. In order to obtain that I use the fact that the gradient of u on $T$ is the rotation by $2\pi/5$ of the gradient of $u$ on the reflection of $T$ w.r.t. the $x$ axis.

I'm pretty sure in this general form the claim is false. You can imagine lots of bodies with rotational symmetries that are non-convex, maybe, and for which a packing with unaligned copies is more efficient. This is not an answer, but more like a hunch.

Thank you very much for performing the computations! Nice answer.

Thank you very much for these computations. I will check them.

@EmilJeřábek: Adding the parameter alpha you modify the $\omega^i$ using some complex numbers. Therefore, the new coefficients $x_i$ would be complex, not real, no? For example if $n=3$, the best bound for the imaginary part is smallest than the one for the real part ($2\sin 2\pi/3$ vs $2$).

I am well aware that $n$ is a valid upper bound for $|z|$, but it's not the best one. I agree that there is not a big improvement over $n$, but I was wondering if the best bounds are known. As you say, for the real and imaginary parts, the computation is straightforward: a sum of cosines or sines.