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. 2 added 1659 characters in body; added 90 characters in body Edit 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 ? 1 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.