FIrst I'd clarify that your notation $e_{ij}^\ell$ actually refers to the matrix with diagonal entries 1, the off-diagonal $(i,j)$ entry equal to $\ell$, and other entries 0.

I don't know what you've read, but
since these matrix calculations are quite old and also deeply embedded in the study of the Congruence Subgroup Problem, it's a good idea to look into some of the relevant older literature. On a concrete level, the emphasis is on the group $\Gamma: = \mathrm{SL}_n(\mathbb{Z})$ and its subgroups of finite index, when $n \geq 3$ (the case $n=2$ being much more complicated). Here the key players are the (normal) *principal congruence* subgroups you've denoted by $\Gamma_n(\ell)$ and the interleaved *elementary* subgroups: inverse images of finite groups generated by elementary/unipotent matrices.

Key references available online include the 1964 announcement by Bass-Lazard-Serre here and the detailed follow-up by Bass-Milnor-Serre here.

Some of the concrete calculations you are looking for are also written down in section 17 of my (typewritten) 1980 Springer Lecture Notes *Arithmetic Groups*.
The older lecture notes *Algebraic K-Theory* by Bass (1968) contain a vast
amount of detail, and of course there are newer treatises including those by Hahn and O'Meara along with a new book by Weibel.

ADDED: The short paper by Tits cited by Aakumadula is definitely helpful for your question, though it's dependent on the earlier work and is not readily available online (nor is the ancient review I wrote). The literature on congruence subgroups is extensive and often far more general than what you need, but I don't see a direct computational proof of the result you read somewhere.
(Advice: Keep track of those sources.) Also, notation varies in the subject, but your choice of $e_{ij}$ is unfortunate since that symbol usually means the matrix with a single nonzero entry $1$. A more usual convention is to write something like $x_{ij}(\ell)$ for your unipotent matrix.