For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward.
Let $g$ have order $p^r$ in $G$ (if not then we are effectively working
in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$
we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond
to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row
(or column if you put the map on the other side) and we see that the image
of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. 
Let $g\in G$. We can write $G=\langle h\rangle\times H$
where $g=p^s h\in H$ and $h$ has order $p^m$ for some
$m\ge s$. We can specify an endomorphism of $G$ by mapping
$h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism
from $H$ to $G$. Then the $h'$ are uniformly distributed
amongst the elements of order $\le p^m$ in $G$ and $g'$ is
mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed
over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow
$p$-subgroups. Then $g\in G$ splits up into its primary components
and each of these behave in the same way, under a random endomorphism,
as in the $p$-group case above.

**Added** I now see that the argument I gave in the prime power case
is valid in the general case too. The key observation is that
a maximal cyclic subgroup of a finite abelian group is a direct summand.
Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup
of order $mn$ containing $\langle g\rangle$. Then the images of
$g$ under random endomorphisms of $G$ are uniformly
distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$
denotes the $mn$-torsion subgroup of $G$.