Let $G$ be a finitely generated, residually finite, non-cohopfian group. Since $G$ is residually finite, we know that there exists a sequence of nested, normal, finite index subgroups $$G = N_0 \rhd N_1 \rhd N_2 \ldots $$ with the trivial intersection.
Question: Is it possible that all $N_i$ have finite abelianization, i.e. the quotient group $N_i \big/ [N_i,N_i]$ is finite for every $i$?
My thoughts so far: It is clear that $G$ can't be finite (finite implies cohopfian) or abelian or free (free groups have infinite abelianization). The closest group I was informed is $D_{\infty} = \; \left<r,s \;|\; srs=r^{-1}, s^2=1 \right>$, which is finitely generated, residually finite and non-cohopfian. However, for any sequence of nested, normal, finite index subgroups with the trivial intersection $D_{\infty} = N_0 \rhd N_1 \rhd N_2 \ldots $, we eventually have $N_i$ contains only rotations, which means $N_i$ is infinite abelian, hence has infinite abelianization.
I think the Grigorchuk group has all of these properties.
It is finitely generated, residually finite and non-cohopfian. Since it is a torsion group and the $N_i$ are all finitely generated, they must have finite abelianization.