Charles, in your answer you're basically discovering the fact that the normalization is not flat: Let $X$ be a non-normal reduced scheme and for simplicity assume that it is of finite type over a field and $\sigma:\widetilde X\to X$ its normalization. Then $\sigma$ is finite. By considering an affine neighbourhood of a non-normal point we may assume that $\sigma$ is a projective morphism. Then the corresponding Hilbert polynomial of the general fiber is $1$ since $\sigma$ is generically an isomorphism so the general fiber is a reduced closed point. However, the fiber over a non-normal point is either not reduced or is reducible and hence its Hilbert polynomial is different than $1$, so $\sigma$ is not flat. Your example is a special case of this.
Perhaps a more interesting question is to ask for a finite non-flat morphism whose target is smooth. Obviously it will not come from an integral element of the fraction field of the target. Also, the the dimension of the players have to be at least $2$, since if the target of a morphism that is dominant on all irreducible components is a smooth curve, then the morphism is automatically flat.
My favorite example of such a morphism is to take two copies of $\mathbb A^d$, $d\geq 2$, intersecting in a single point and mapping them to $\mathbb A^d$ in the obvious way. The target is smooth, the morphism is finite, it is étale outside a single point, but it is not flat. You can try to prove this directly or to use the fact that a finite morphism whose target is smooth is flat if and only if the source of the morphism is Cohen-Macaulay. It is really easy to see that two copies of $\mathbb A^d$, $d\geq 2$, intersecting in a single point is not Cohen-Macaulay.