This is for a non-Noetherian $A$ right? Because if $A$ is Noetherian and $M$ is finite, $\operatorname{Hom}_A(M,A)=M^\vee$ is always finite.

Since you mention Seidenberg's paper, I imagine you are interested in the non-Noetherian case. However, when $A$ is Noetherian, $\dim A \le \dim A[It] \le \dim A + 1$, with equality on the left $I$ is contained in a minimal prime of maximal dimension ($\dim A/\mathfrak{p} = \dim A$) and equality on the right if $I$ is not contained in such a prime. This is Swanson-Huneke, "Integral Closure of Ideals, Rings, and Modules", Theorem 5.1.4. They do not mention the non-Noetherian case in that subsection.

Of course, given one example of a graded algebra $A$, one also has all of its Veronese subrings $A^{(d)}$, whose $k$th graded piece is $[A]_{kd}$, e.g. $\mathbb{Z}[T]^{(4)} = \mathbb{Z}[T^4]$. This example is isomorphic to $\mathbb{Z}[T]$, but in general Veronese subrings are not isomorphic to their base ring.

To follow up; if $R$ is a henselian domain, then by stacks.math.columbia.edu/tag/0C24, we have that $\overline{R}$ is a local ring since the number of minimal primes of $R^h$ is the same as the number of maximal ideals of $\overline{R}$, correct?

I think this question is much more suited to MathStackExchange than MathOverflow. But, to answer your question, for the case that $\operatorname{ht}(\mathfrak{p}) = 1$, use that $R$ is a UFD so height one primes are principal, meaning $\mathfrak{p}=(x)$ for some non-zerodivisor $x \in R$. For the case that $\operatorname{ht}(\mathfrak{p})=n-1$, as @JasonStarr says, since $\mathfrak{p}$ is prime, $R/\mathfrak{p}$ is a one-dimensional domain (hence reduced), and thus any nonzero element is a non-zerodivisor so $R$ is Cohen-Macaulay.

I believe this MSE question helps with the (finite) group case. math.stackexchange.com/questions/1599089/…

(I liked both answers, but accepted Richard Stanley's because I was looking for references and he included a reference.)

Thank you! I had looked at your book and several others when thinking about Ehrhart polynomials recently. In fact, your paper about combinatorial reciprocity inspired my initial thoughts about this question!

Very nice! Thank you for the answer. I suppose I should have expected a canonical module would show up. I am most interested in non-Cohen-Macaulay examples, but this gives me somewhere to start.

@DaveBenson Yes, I was thinking of it asymptotically, so "almost" is good enough for me.

@DaveBenson Thank you for the reference! I believe this is called "quasi-polynomial" behavior, isn't that right? It is very interesting to know that there are some negative zeroes of $h_R(n)$. Is there any, say topological, interpretation about the Lie group that comes about from knowing the zeroes of the Hilbert polynomial of these rings?