I'll show that if $F$ is a free group of rank at least three, then $\mathbb{Z}F$ has a finitely generated non-Hopfian module. (So this is not an answer to the question, but it seems relevant and is too long for a comment.)
Suppose $R$ is any ring with a finitely generated non-Hopfian right module $M$.
Then the ring $M_2(R)$ of $2\times2$ matrices over $R$ is another example, because it's Morita equivalent to $R$ (or, more explicitly, the $M_2(R)$-module $(M\text{ }M)$ of row vectors with entries from $M$ is a finitely generated non-Hopfian $M_2(R)$-module).
Also $M_2(R)$ is generated as a ring by units: If $X=\{x_i\vert i\in I\}$ is a generating set for $R$ in the strong sense that there is not even a proper nonunital subring of $R$ containing $X$, then it is easy to check that
$$Y=\left\{\pmatrix{x_i&1\\1&0}\middle\vert i\in I\right\}\cup\left\{\pmatrix{0&1\\1&0}\right\}$$
is a generating set for $M_2(R)$ consisting of units.
So if $F$ is the free group on the set $Y$, there is a natural surjective ring homomorphism $\mathbb{Z}F\to M_2(R)$, and so we can regard $(M\text{ }M)$ as a finitely generated non-Hopfian right $\mathbb{Z}F$-module.
If we take $R$ to be the ring
$$R=\mathbb{Z}\left\langle x,y\middle\vert xy=1\right\rangle,$$
then the maps given by left multiplication by $y$ and $x$ exhibit the free right module $R$ as a proper direct summand of itself, and so $R$ is a non-Hopfian module for itself. Since $R$ is generated by two elements, $M_2(R)$ is generated by three units, and so is a quotient of $\mathbb{Z}F$ for $F$ free of rank three.