This is not a true general fact for any abelian category, as Vakil points out in 1.6.12. He gives the following counterexample, which puzzled for it is in the category of abelian groups, and every abelian group is a module over $\mathbb Z$. So my question is: why is his counterexample not also a counterexample in the category $\mathcal{Mod}_\mathbb Z$?
⋆⋆ Here is a counterexample. Because the axioms of abelian categories are self-dual, it suffices to give an example in which a filtered limit fails to be exact, and we do this. Fix a prime $p$. In the category $\mathcal{Ab}$ of abelian groups, for each positive integer $n$, we have an exact sequence $\mathbb Z \rightarrow \mathbb Z/(p^n)\rightarrow 0$. Taking the limit over all n in the obvious way, we obtain $\mathbb Z\rightarrow \mathbb Z_p\rightarrow 0$, which is certainly not exact.) **
The category $\text{Mod-}A$ of right modules over a ring is known to be a Grothendieck category; in particular, filtered colimits are exact in this category.
This category is not “coGrothendieck”, though. Indeed it's not hard to show that if an abelian category is both Grothendieck and coGrothendieck, then it is the one object and one morphism category.
The self-dual nature of the axioms for an abelian category doesn't allow to conclude that for any property of an abelian category $\mathcal{A}$ then the dual property holds in $\mathcal{A}$. Exactness of filtered colimits in $\text{Mod-}A$ is an example.
What can be derived from the self-duality of the axioms is that if a property holds in all abelian categories, then also its dual property holds in all abelian categories. So the “counterexample” you present only shows that in some abelian categories filtered colimits are not exact (for instance in the opposite category of $\text{Mod-}\mathbb{Z}$).