What sort of ring is the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$?

This is just the ring of algebraic integers, which is a Bézout domain. That means that every finitely generated ideal is principal. As there do exist non-finitely generated ideals which are not principal, it is not a Noetherian domain.

For example, consider the ideal generated by $\{2^{1/n}\colon n\ge1\}$. As no generating set is contained in any finite extension of $\mathbb{Q}$, it is not finitely generated. On the other hand, any quotient by a nonzero prime ideal $\mathfrak{p}$ gives an algebraic extension of $\mathbb{Z}/\mathfrak{p}\cap\mathbb{Z}$, so is a field, and $\mathfrak{p}$ is maximal. So it does have Krull dimension 1.

The other common example of a Bézout domian which is not Noetherian is the ring of entire functions on the complex plane.

No, it is not noetherian. The ideals generated by $2^{1/2^n}, n = 1, 2, \dots$ form a strictly ascending chain that does not stabilize. It is of Krull dimension one (every nonzero prime ideal is maximal) by lying over and going up.

To your related question "For example is flat?" the answer is yes. The point is that over a Dedekind domain, more generally over a Prüfer ring, a module is flat if and only if it is torsion free( see here). On the other hand I'm not so sure if I can say more about of the morphism. It is not going to be locally finite, locally of finite type, locally Noetherian etc...