$L$ is regular. I'll prove this for $n=1$; then it follows for arbitrary $n$, as the intersection of regular languages is itself regular.
Define
$$L = \{x \in \Sigma^* \mid \{x\} \cdot A \subseteq B\}.$$
I will construct a finite-state automaton $M_L$ for $L$, proving that $L$ is regular.
Let $M_A,M_B$ be deterministic finite-state automata for the languages $A,B$. WLOG we can assume they have no useless or unreachable states. Let $s_0,t_0$ denote the start states for $M_A,M_B$.
$M_L$'s states have the form $\langle s,t \rangle$ or $\langle \bot,t \rangle$, where $s,t$ range over states of $M_A,M_B$, respectively.
$M_L$ has a transition $\langle s,t \rangle \stackrel{c}{\to} \langle s',t' \rangle$ if $M_A$ has a transition $s \stackrel{c}{\to} s'$ and $M_B$ has a transition $t \stackrel{c}{\to} t'$. Also, $M_L$ has a transition $\langle \bot,t \rangle \stackrel{\epsilon}{\to} \langle s_0,t \rangle$ for each $t$.
Mark the state $\langle s_0,t \rangle $ as accepting in $M_L$ iff for every path $\langle s_0,t \rangle \leadsto \langle s',t' \rangle$ in $M_L$ such that $s'$ is accepting in $M_A$, $t'$ is accepting in $M_B$. (*)
You should be able to prove that $M_L$ forms a finite-state automaton for $L$. I'll let you work out the details of the proof. If you want, you can easily adjust $M_L$ to be deterministic by removing the few $\epsilon$ transitions. This gives you the result you desired, as well as a constructive algorithm to form an automaton recognizing the language. Combine this with the standard method (product automaton) for the intersection of multiple regular languages, and you obtain a solution for arbitrary $n$ as well.
Footnote (*): Also, you can efficiently find all such states using a single depth-first search. Let
$$I = \{\langle s',t' \rangle \mid s' \text{ is accepting}, t' \text{ is not accepting}\}.$$
Now do a depth-first search backwards through $M_L$, starting from the set of states $I$ defined as follows, to find all states $R$ that are reachable (going backwards) from $I$. Mark every state of the form $\langle s_0,t\rangle$ that is not in $R$ as accepting in $M_L$; all other states of $M_L$ are not accepting.
