Should we expect Jahn–Teller distortion in bis(ethylenediamine)nickel(II)?
There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a single electron with parallel spins leads to no degeneracy problems. An additional electron will mean that one of the two $\mathrm{e_g^*}$ orbitals is doubly occupied while the other is singly occupied—but this is a degenerate state as the energy is the same whether it is $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ that is doubly occupied.

This degeneracy of possible states in copper(II)—or any other $\mathrm d^9$ system—is the underlying reason for Jahn-Teller distortions. By lowering symmetry from $O_\mathrm h$ to $D_\mathrm{4h}$, these two orbitals are no longer degenerate and they can (and must) diverge in energy. Thus, you have a clear ‘winner’—the lower-energy orbital which receives an electron pair—and a ‘loser’—the higher-energy orbital which receives the single electron. Looking only at $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ it doesn’t matter whether the symmetry reduction is performed by moving the $z$-ligands further away or closer but it does make a difference for the other three d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$) for which lengthening the $\ce{M-L}$ bonds along the $z$ axis proves favourable overall.

A $\mathrm d^8$ system at first glance has nothing to gain by this distortion. Rather than having a stabilisation that benefits two electrons at the expense of one, it can only stabilise one electron at the expense of one. This is energetically not favourable as there is nothing to be gained (but probably something to be lost) so a ‘simple’ Jahn-Teller distortion does not occur; nickel(II) prefers to remain octahedral in $\ce{[ML6]}$ systems.

Note however that there is one possibility when such a stabilisation is beneficial: if the ‘distortion’ is of such a large magnitude that the energy difference between $\mathrm d_{x^2-y^2}$ and the next-lowest orbital is larger than the spin-pairing energy. In this case, we could say that nickel(II) drops from a high-spin state to a low-spin state, although these terms are not usually used in this context. Geometrically, this ‘distortion’ is so significant that we term the resulting complex square planar rather than distorted octahedron; in general there is no longer any evidence of coordination bonding along the $z$ axis. A number of nickel(II) complexes with strong mid-field or strong-field ligands are indeed square planar for this reason such as $\ce{[Ni(dmg)2]}$, the dimethylglyoxime (or diacetyldioxime) complex.