Complexity of DFA intersection in this specific case? In general, the size of the DFA that recognizes the intersection of $n$ languages is exponential in $n$. However, in my case I am computing the intersection of a very restricted subset of possible languages. I am curious if this weakens the lower complexity bound.

With an alphabet $\Sigma$, for the $i$th language, I have set $A_i \subseteq \Sigma$ and another set $B_i \subseteq \Sigma$. The $i$th language $L_i$ consists of strings containing any number of symbols in $A_i$, followed by one symbol in $B_i$, followed by any number of symbols in $\Sigma$. I would like to find a DFA that recognizes $L = L_1 \cap L_2 \cap ... \cap L_n$.

Do these restrictions on the input languages eliminate the exponential lower bound?
The precise bound is $2^n$. The lower bound was given in the comments: the state complexity of $A^*a_1A^* \cap \dotsm \cap A^*a_nA^*$ is $2^n$.
For the upper bound, it suffices to observe that if $B$ and $C$ are subsets of the alphabet $A$, then the language $B^*CA^* = (B - C)^*CA^*$ is recognised by a 2-state DFA. It follows that the complexity of the intersection of $n$ such languages is upper bounded by $2^n$.