Fix an $A$ with CM by $K$, and for each $D$, let $A_D$ be the quadratic twist of $A$ by $K(\sqrt{D})/K$. Also let $$h_{sf}(D)=\min(h(Dd^2):d\in K^*)$$ denote the "square-free height" of $D$. Then 
$$
  h_{\text{Faltings}}(A_D) \gg h_{sf}(D),
$$
which shows that there is no lower bound of the sort that you want, since $K$ is fixed, while $h_{sf}(D)$ can be arbitrarily large.

Or do you mean to take the semi-stable Faltings height, i.e., the height obtained after going to a field where $A$ has semi-stable reduction. For CM abelian varieties, this would be a field where $A$ has everywhere good reduction, so the Faltings height comes entirely from the archimedean places. In this case, you can use the fact that the Faltings height is more-or-less equal to the height of the associated point in moduli space. (At least, equal enough to talk about boundedness.) For a principally polarized CM abelian variety, the moduli point is essentially given by the periods, which are more-or-less a basis for $\mathcal{O}_K$ over $\mathbb{Z}$. So it seems that one might well be able to get a bound in terms of $\hbox{Disc}(K)$.

You might try looking first at the case of elliptic curves, where the relation between the Faltings height and the periods is very explicit.