Let $X\longrightarrow X'$ be a smooth proper map of smooth proper schemes defined over $\mathbb{Z}[1/S]$, where $S$ is a finite set of primes. Assume $X'$ is a curve of positive genus, and $X$ is a surface, which is in fact a relative curve of positive genus.
Is it possible to say anything meaningful about the dimension of $H^4_{\text{ėt}}(X, \mathbb{Q}_p)$?
Are you absolutely sure you want to compute $p$-adic etale cohomology for a smooth proper $\mathbb{Z}[1/S]$-scheme with $p \notin S$, so $p$ is not invertible on $X$? This will be painful, and I strongly suspect it isn't what you really want.
If you are willing to take $p \in S$ then the situation becomes much better. Then you have a spectral sequence converging to $H^*_{et}(X, \mathbb{Q}_p)$ with $E_2$-page $H^j(\mathbb{Z}[1/S], H^i(X_{\overline{\mathbb{Q}}}, \mathbb{Q}_p))$. This can only be non-zero for $j = 0, 1, 2$. Moreover, the contribution from $H^4(X_{\overline{\mathbb{Q}}}, \mathbb{Q}_p)^{G_\mathbb{Q}}$ is zero, since $H^4(X_{\overline{\mathbb{Q}}}, \mathbb{Q}_p) \cong \mathbb{Q}_p(-2)$ as a Galois module by Poincare duality.
So you are left with contributions from $H^2(H^2)$ and $H^1(H^3)$. The $H^2(H^2)$ term is almost certainly zero by Jannsen's conjecture (see e.g. here), although verifying this unconditionally is super-hard. So you are left with the $H^1(H^3)$ term. It should be possible to get some information about this out of Tate's Euler characteristic formula (combined, again, with Jannsen's conjecture to get vanishing of the $H^2$). Back-of-the-envelope computations suggest that the answer should be roughly the sum of the genus of $X'$ and the relative genus of $X / X'$.
Note that the question would become much more difficult if you asked about Tate twists such as $H^4(X, \mathbb{Q}_p(2))$, lying in the range where Jannsen's conjecture doesn't apply. "Here be dragons" (or, rather, here be orders of vanishing of L-functions...)
