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For (Remove generalization to non-noetherian rings).

Remark. Let $A$ be any finite dimensional noetherian ring. If $A$ having has a finite residue field, there exists $q$ such that $e_A(n)$ coincides asymptotically with $ \lceil \log_q n \rceil $.

Proof. Let $q$ be the smallest cardinality of the residue fields of $A$. Then $f_A(n)=e_{\mathbb F_q}(n)$. It is enough to show that $e_A(n)\le e_{\mathbb F_q}(n)$ take for $q$ big enough because $f_A(n)\le e_A(n)$.

Let $A_0\to A$ be a homomorphism with $A_0$ a finitely generated $\mathbb Z$-algebra. Then $A_0$ has finite Krull dimension $d_0$ and $$e_A(n)\le e_{A_0}(n)\le \max \{ d_0+1, f_{A_0}(n)\}.$$ There are only finitely many maximal ideals of $A_0$ with residue fields $k$ of cardinality smaller than $q$ (see e.g. the above preprint, Lemma 1.11). Now we enlarge $A_0$ to make $f_{A_0}(n)\le f_A(n)$. If $k_0$ is a subfield of a residue field $k$ of $A$, we can enlarge $A_0$ so that these kind smallest cardinality of residue fields are either equal to some the finite residue fields $k$ of $A$ (when $k$ is finite) or has more than $q$ elements (if $k$ is infinite). A$. If a finite residue field $k_0$ of characteristic $\ell \le q-1$ is not a subfield of a residue field of $A$, then $\ell$ is invertible in $A$, and we can replace $A_0$ with $A_0[1/\ell]$. We finally get an $A_0$ whose finite residue fields all have cardinality $\ge q$. Hence $f_{A_0}(n)\le f_A(n)$ for $n$ big enough (such that $\log_q n$ is bigger then $d_0+1$).

Remark. I don't know what happens if all residue fields of $A$ are infinite. For any $q\ge 2$, the above reasonning shows that asymptically $e_A(n)$ is smaller or equal to $\lceil \log_q n \rceil$. If $A$ is noetherian of finite dimension, then $e_A(n)$ is bounded hence asymptotically constant (because it is increasing). But is it true in general ? It would be interesting to decide whether these properties hold without noetherian and finite-dimensional hypothesis.
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For any $A$ having a finite residue field, there exists $q$ such that $e_A(n)$ coincides asymptotically with $ \lceil \log_q n \rceil $.

Proof. Let $q$ be the smallest cardinality of the residue fields of $A$. Then $f_A(n)=e_{\mathbb F_q}(n)$. It is enough to show that $e_A(n)\le e_{\mathbb F_q}(n)$ for $q$ big enough because $f_A(n)\le e_A(n)$.

Let $A_0\to A$ be a homomorphism with $A_0$ a finitely generated $\mathbb Z$-algebra. Then $A_0$ has finite Krull dimension $d_0$ and $$e_A(n)\le e_{A_0}(n)\le \max \{ d_0+1, f_{A_0}(n)\}.$$ There are only finitely many maximal ideals of $A_0$ with residue fields $k$ of cardinality smaller than $q$ (see e.g. the above preprint, Lemma 1.11). Now we enlarge $A_0$ to make $f_{A_0}(n)\le f_A(n)$. If $k_0$ is a subfield of a residue field $k$ of $A$, we can enlarge $A_0$ so that these kind of residue fields are either equal to some residue fields $k$ of $A$ (when $k$ is finite) or has more than $q$ elements (if $k$ is infinite). If a finite residue field $k_0$ of characteristic $\ell \le q-1$ is not a subfield of a residue field of $A$, then $\ell$ is invertible in $A$, and we can replace $A_0$ with $A_0[1/\ell]$. We finally get an $A_0$ whose finite residue fields all have cardinality $\ge q$. Hence $f_{A_0}(n)\le f_A(n)$ for $n$ big enough (such that $\log_q n$ is bigger then $d_0+1$).

Remark. I don't know what happens if all residue fields of $A$ are infinite. For any $q\ge 2$, the above reasonning shows that asymptically $e_A(n)$ is smaller or equal to $\lceil \log_q n \rceil$. If $A$ is noetherian of finite dimension, then $e_A(n)$ is bounded hence asymptotically constant (because it is increasing). But is it true in general ?

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You already completely solved the question over fields, noetherian artinian rings and $\mathbb Z$.

Let $A$ be any commutative unitary ring. Then the maximum $f_A(n)$ of all $e_k(n)$ when $k$ runs the residue fields of $A$ (at maximal ideals) satisfies clearly $e_A(n)\ge f_A(n)$ by your (4).

Suppose $A$ is noetherian of dimension $d$, then $$f_A(n) \le e_A(n) \le \max \{ d+1, f_A(n)\}.$$

Proof. Let $m=\max \{ d+1, f_A(n)\}$. We want to show that the affine space $\mathbb A^m$ over $A$ contains $n$ disjoint sections. Let $r\le n-1$ be such that $\mathbb A^m$ contains $r$ disjoint sections. We are going to show that $\mathbb A^m$ contains one more section disjoint from the previous one. This will prove the claim.

Let $T$ be the union of $r$ sections. For every residue field $k$ of $A$, $\mathbb A^m_k$ contains at least $r+1$ rational points. In particular, $T$ doesn't contain $\mathbb A^m_k(k)$. By hypothesis, we also have $\dim T=\dim A< m$. By Proposition 1.10 of this preprint, there is a section in $\mathbb A^m$ disjoint from $T$ and we are done.