More precisely, is every abelian group a colimit $\text{colim}_{j \in J} F(j)$ over a diagram $F : J \to \text{Ab}$ where each $F(j)$ is isomorphic to $\mathbb{Z}$?

Note that this does not follow from the statement that every abelian group has a presentation, which is equivalent to the statement that every abelian group is a coequalizer of a pair of maps between free abelian groups, hence every abelian group is an *iterated* colimit of copies of $\mathbb{Z}$. A single colimit $A = \text{colim}_{j \in J} F(j)$ of copies of $\mathbb{Z}$ is in particular the coequalizer of a pair of maps between free abelian groups, but the maps have a very special form, which works out explicitly to imposing the following constraint:

$A$ must have a presentation by generators and relations in which the only relations say that some generator is a multiple of some other generator.

Examples of abelian groups admitting a presentation of this form include cyclic groups and localizations of $\mathbb{Z}$, and the class of all such groups is closed under coproducts.

But I see no reason to believe that every abelian group admits a presentation of this form, and in particular I believe that the $p$-adic integers doesn't. Tyler Lawson sketched a proof of this in the homotopy theory chat but it had a gap; the subsequent discussion may have filled the gap but I didn't follow it, and in any case I'd like someone to write up the details.

Mike Shulman wrote a lovely note about various different senses in which an object or objects of a category can generate it; in the terminology of that note, the question is whether $\mathbb{Z}$ is *colimit-dense* in $\text{Ab}$. Until a week or so ago, if you had asked me, I would have answered without hesitation that $R$ is colimit-dense in $\text{Mod}(R)$, and I doubt that I was alone in this...

mightbe a counterexample is: the direct product of an infinite number of copies of Z $\endgroup$ – Yemon Choi May 5 '15 at 17:29