Let $\Sigma$ be a finite alphabet. All strings below are over $\Sigma$.
Definitions:
If a string $s = vw$, then $v$ is a $\textit{prefix}$ of $s$ and $w$ is a $\textit{suffix}$ of $s$.
For a language $L \subseteq \Sigma^*$, define $L_n \subseteq L$ to be the set of strings in $L$ of length $n$: $L_n = \{s \mid s \in L \; \textrm{and} \; |s| = n \}$.
Since $L_n$ is finite, it can be recognized by a deterministic finite-state automaton (DFA) $M_{L,n}$ that is minimal and acyclic.
A $\textit{level}$ of an acyclic DFA is the set of all states at the same depth as measured from the start state.
We are given the following constraint on $L$:
$$ | L_n | = O(n^c) $$ where $c$ is a constant that depends on $L$ but not on $n$.
Now if we take the set of all prefixes of a certain length $d$,
$$ S_{n,d} = \{ v \mid vw \in L_n, |v| = d \} $$
$S_{n,d}$ is also similarly bounded:
$$ | S_{n,d} | = O(n^c) $$
So even if, in the worst case, all strings in the $S_{n,d}$ are Myhill-Nerode distinguishable from each other, there are therefore only $O(n^c)$ states at depth $d$ in $M_{L,n}$.
So can we conclude that: given a language $L$ such that $ | L_n | = O(n^c) $ where $c$ is a constant, the minimal acyclic DFA that recognizes the subset $L_n \subseteq L$ has only $O(n^{k})$ states for some fixed $k$?