I think $\mathbb{P}^n_{\mathbb{F}_p}$ should have $p^n + \cdots + p + 1 = \frac{p^{n+1} - 1}{p - 1}$ points (e.g. $\mathbb{P}^1_{\mathbb{F}_p}$ should have $p$ ``finite'' points plus a ''point at infinity'').

If $X$ and $\mathbb{P}^n$ ae birationally equivalent I don't think you can get an inequality either way. For example $X = \mathbb{P}^1 \times \mathbb{P}^1$ has $(p+1)^2 = p^2 + 2 p + 1$ points, larger than $p^2 + p + 1$ points on $\mathbb{P}^2$. On the other hand, the nodal curve $y^2 = x^2(x+1)$ has $p$ points but it birational to $\mathbb{P}^1$.

If you have a finite type scheme $X$ over $\mathrm{Spec}(\mathbb{Z})$ such that there is a dominant morphism $\mathbb{P}^n \to X_{\mathbb{Q}}$ then by spreading out you get a dense open $U \subset \mathrm{Spec}(\mathbb{Z})$ such that $\mathbb{P}^n \times U \to X_U$ is dominant on each fiber. Therefore this holds for all but finitely many primes $p$.

If you really don't want a morphism only a rational map, I think the argument still works but you replace $\mathbb{P}^n_{\mathbb{Z}}$ with some open containing the locus in the generic fiber where your rational map is defined then you do spreading out for this scheme.

Here are some examples showing that ''for all but finitely many $p$'' is really necessary. You can do something silly like make $X$ be some weird variety only supported over $\mathbb{F}_p$ for some fixed $p$. However, you probably also want to impose that $X \to \mathrm{Spec}(\mathbb{Z})$ is flat (or at least flat over the set of primes you care about).

Consider $X = V(xy + p z^2) \subset \mathbb{P}^2_{\mathbb{Z}} = \mathrm{Proj}(\mathbb{Z}[x,y,z])$. Then the generic fiber is a smooth conic with a rational point $[p,-1,1]$ and thus isomorphic to $\mathbb{P}^1$. However, the fiber over $p$ is the union of two lines and thus cannot be dominated by $\mathbb{P}^1$.

I think $\mathbb{P}^n_{\mathbb{F}_p}$ should have $p^n + \cdots + p + 1 = \frac{p^{n+1} - 1}{p - 1}$ points (e.g. $\mathbb{P}^1_{\mathbb{F}_p}$ should have $p$ ``finite'' points plus a ''point at infinity'').

If $X$ and $\mathbb{P}^n$ are birationally equivalent I don't think you can get an inequality either way. For example $X = \mathbb{P}^1 \times \mathbb{P}^1$ has $(p+1)^2 = p^2 + 2 p + 1$ points, larger than $p^2 + p + 1$ points on $\mathbb{P}^2$. On the other hand, the nodal curve $y^2 = x^2(x+1)$ has $p$ points but it birational to $\mathbb{P}^1$.

If you have a finite type scheme $X$ over $\mathrm{Spec}(\mathbb{Z})$ such that there is a dominant morphism $\mathbb{P}^n \to X_{\mathbb{Q}}$ then by spreading out you get a dense open $U \subset \mathrm{Spec}(\mathbb{Z})$ such that $\mathbb{P}^n \times U \to X_U$ is dominant on each fiber. Therefore this holds for all but finitely many primes $p$.

If you really don't want a morphism only a rational map, I think the argument still works but you replace $\mathbb{P}^n_{\mathbb{Z}}$ with some open containing the locus in the generic fiber where your rational map is defined then you do spreading out for this scheme.

Here are some examples showing that ''for all but finitely many $p$'' is really necessary. You can do something silly like make $X$ be some weird variety only supported over $\mathbb{F}_p$ for some fixed $p$. However, you probably also want to impose that $X \to \mathrm{Spec}(\mathbb{Z})$ is flat (or at least flat over the set of primes you care about).

Consider $X = V(xy + p z^2) \subset \mathbb{P}^2_{\mathbb{Z}} = \mathrm{Proj}(\mathbb{Z}[x,y,z])$. Then the generic fiber is a smooth conic with a rational point $[p,-1,1]$ and thus isomorphic to $\mathbb{P}^1$. However, the fiber over $p$ is the union of two lines and thus cannot be dominated by $\mathbb{P}^1$.