Why should $\lim_{V\to\infty} \frac{1}{V} \ln Q(z, V, T)$ have a finite limit? In the book Intro. Statistical Physics by K.Huang, on page 174, it is given that

In the thermodynamic limit $V \rightarrow \infty,$ we expect that:
$$
\frac{1}{V} \ln Q(z, V, T) \underset{V \rightarrow \infty}{\longrightarrow} \text { Finite limit. }
$$
where Q is the grand canonical partition function.

This is expected but is there any mathematical or physical reason and/or evidence/explanation for why this is/should be the case?
 A: There is no mathematical proof just because, in general, it is not true that the limit exists or it is finite. Of course, we would expect a finite limit as a precondition for a thermodynamic interpretation of the statistical mechanics formula.
The right question is not about the reason for a finite limit, but to ask the question do we have a good characterization of the Hamiltonians which ensure the existence of thermodynamic limit?
Indeed, a set of sufficient conditions for the existence of the thermodynamic limit, ensuring at the same time the correct properties of convexity of the resulting fundamental equation, is known for different classes of systems. For an overview see this paper.
A: In grand canonical ensemble, the partition function $Q$ and the grand potential has relation
$$
\Phi = -K T \ln Q.
$$
The grand potential (also known as Landau free energy) in thermodynamics can be derived using Legendre transformation from internal energy $U(N,V, S)$, where $S$ is the entropy:
The Helmholtz free energy
$$
F = F - TS; \text{  and  } F = F(N, V, T).
$$
Then, the grand potential:
$$
\Phi = F - N\mu; \text{  and  } \Phi = \Phi(\mu, V, T).
$$
The grand potential is an extended quantity, as well the volume $V$, but $T$ and $\mu$ are intensive quantity. The form of grand potential:
$$
\Phi = P V.
$$
Thus,
$$
\lim_{\text{thermal-limit}} \frac{\ln Q(\mu,V, T)}{V} = -\frac{P}{KT}.
$$
$P$ and $T$ are intensity variables will not change as the system becomes larger.
