This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem

 

Let $f:X\to Y $ be a dominant morphism of finite type schemes over $\mathbb{Z}$ with $X_{\mathbb{Q}}, Y_{\mathbb{Q}}$ geometrically integral of dimensions $n,m$, respectively. Then there exists an absolute constant $c$ such that $$ \#f(X(\mathbb{F}_p)) \geq c p^{m} + O(p^{m-1/2})$$ for all sufficiently large primes $p$, where the implied constant in the big-$O$ is independent of $p$.

To prove this, we use the following key lemma.

Lemma

 

There exists an absolute constant $c'$ such that for all $y \in Y(\mathbb{F}_p)$ we have $$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)} $$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $c'$).

Proof of the Theorem: There exists a proper closed subset $Z \subset Y$ such that every fibre outside of $Z$ has dimension $n-m$ (this follows from Lemma 36.28.1. of https://stacks.math.columbia.edu/tag/05F6 applied to $f: X \to Y$). Moreover the number of $\mathbb{F}_p$-points in $Z$ is at most $c''p^{m-1}$ for some absolute constant $c''$. Thus replacing $Y$ by $Y \setminus Z$, we may assume that the dimension of every fibre of $f$ is exactly $n-m$. Thus by the key lemma, we have $$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p)) $$ However, by the Lang-Weil estimates, we have $$\# X(\mathbb{F}_p) = p^n + O(p^{n-1/2})$$ as $p \to \infty$, since $X_{\mathbb{Q}}$ is geometrically integral. The result now easily follows with $c = 1/c'$. $\Box$

Remark: The constant $c$ depends on the number of irreducible components of the generic fibre of $f$ over the algebraic closure. In particular, if the generic fibre of $f$ is geometrically integral, then you can take $c = 1$.

This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem

Let $f:X\to Y $ be a dominant morphism of finite type schemes over $\mathbb{Z}$ with $X_{\mathbb{Q}}, Y_{\mathbb{Q}}$ geometrically integral of dimensions $n,m$, respectively. Then there exists an absolute constant $c$ such that $$ \#f(X(\mathbb{F}_p)) \geq c p^{m} + O(p^{m-1/2})$$ for all sufficiently large primes $p$, where the implied constant in the big-$O$ is independent of $p$.

To prove this, we use the following key lemma.

Lemma

There exists an absolute constant $c'$ such that for all $y \in Y(\mathbb{F}_p)$ we have $$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)} $$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $c'$).

Proof of the Theorem: There exists a proper closed subset $Z \subset Y$ such that every fibre outside of $Z$ has dimension $n-m$ (this follows from Lemma 36.28.1. of https://stacks.math.columbia.edu/tag/05F6 applied to $f: X \to Y$). Moreover the number of $\mathbb{F}_p$-points in $Z$ is at most $c''p^{m-1}$ for some absolute constant $c''$. Thus replacing $Y$ by $Y \setminus Z$, we may assume that the dimension of every fibre of $f$ is exactly $n-m$. Thus by the key lemma, we have $$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p)) $$ However, by the Lang-Weil estimates, we have $$\# X(\mathbb{F}_p) = p^n + O(p^{n-1/2})$$ as $p \to \infty$, since $X_{\mathbb{Q}}$ is geometrically integral. The result now easily follows with $c = 1/c'$. $\Box$

Remark: The constant $c$ depends on the number of irreducible components of the generic fibre of $f$ over the algebraic closure. In particular, if the generic fibre of $f$ is geometrically integral, then you can take $c = 1$.

This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem

Let $f:X\to Y $ be a dominant morphism of finite type schemes over $\mathbb{Z}$ with $X_{\mathbb{Q}}, Y_{\mathbb{Q}}$ geometrically integral of dimensions $n,m$, respectively. Then there exists an absolute constant $c$ such that $$ \#f(X(\mathbb{F}_p) \geq c p^{m}$$ for all sufficiently large primes $p$.

To prove this, we use the following key lemma.

Lemma

There exists an absolute constant $c'$ such that for all $y \in Y(\mathbb{F}_p)$ we have $$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)} $$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $c$).

Proof of the Theorem: There exists a proper closed subset $Z \subset Y$ such that the every fibre outside of $Z$ has dimension $n-m$. Moreover the number of points in $Z$ is at most $c''p^{m-1}$ for some absolute constant $c''$. Thus replacing $Y$ by $Y \setminus Z$, we may assume that the dimension of every fibre of $f$ is exactly $n-m$. Thus by the key lemma, we have $$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p)) $$ However, by the Lang-Weil estimates, we have $$\# X(\mathbb{F}_p) \sim p^n$$ as $p \to \infty$. The result now easily follows. $\Box$

Remark: The constant $c$ depends on the number of irreducible components of the generic fibre over the algebraic closure. In particular, if the generic fibre is geometrically integral, then you can take $c = 1$.

This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem

Let $f:X\to Y $ be a dominant morphism of finite type schemes over $\mathbb{Z}$ with $X_{\mathbb{Q}}, Y_{\mathbb{Q}}$ geometrically integral of dimensions $n,m$, respectively. Then there exists an absolute constant $c$ such that $$ \#f(X(\mathbb{F}_p)) \geq c p^{m} + O(p^{m-1/2})$$ for all sufficiently large primes $p$, where the implied constant in the big-$O$ is independent of $p$.

To prove this, we use the following key lemma.

Lemma

There exists an absolute constant $c'$ such that for all $y \in Y(\mathbb{F}_p)$ we have $$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)} $$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $c'$).

Proof of the Theorem: There exists a proper closed subset $Z \subset Y$ such that every fibre outside of $Z$ has dimension $n-m$ (this follows from Lemma 36.28.1. of https://stacks.math.columbia.edu/tag/05F6 applied to $f: X \to Y$). Moreover the number of $\mathbb{F}_p$-points in $Z$ is at most $c''p^{m-1}$ for some absolute constant $c''$. Thus replacing $Y$ by $Y \setminus Z$, we may assume that the dimension of every fibre of $f$ is exactly $n-m$. Thus by the key lemma, we have $$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p)) $$ However, by the Lang-Weil estimates, we have $$\# X(\mathbb{F}_p) = p^n + O(p^{n-1/2})$$ as $p \to \infty$, since $X_{\mathbb{Q}}$ is geometrically integral. The result now easily follows with $c = 1/c'$. $\Box$

Remark: The constant $c$ depends on the number of irreducible components of the generic fibre of $f$ over the algebraic closure. In particular, if the generic fibre of $f$ is geometrically integral, then you can take $c = 1$.