The suggestions in the comment by usul are correct.

Indeed, let
\begin{equation}
p:=P(B),\quad B:=\bigcap_1^n A_j,\quad p_j:=P(A_j).
\end{equation}
We want to find the extreme values of
\begin{equation}
d:=p-\prod_1^n p_j.
\end{equation}

Clearly, $p_j\ge p$ for all $j$ and hence
\begin{equation}
d\le p-p^n\le\max_{0\le p\le1}(p-p^n)=r-r^n,
\end{equation}
where $r:=1/n^{1/(n-1)}$. On the other hand, if $A_1=\dotsb=A_n$ and $p_j=r$ for all $j$, then $d=r-r^n$. So,
\begin{equation}
\max d=r-r^n=\frac{n-1}{n^{n/(n-1)}}
\end{equation}
(so that $\max d\to1$ as $n\to\infty$).

Next,
\begin{equation}
B^c=\bigcup_1^n A_j^c,
\end{equation}
where $^c$ denotes the complement. So,
\begin{equation}
1-p=P(B^c)\le\sum_1^n P(A_j^c)=n-\sum_1^n p_j,
\end{equation}
so that $\sum_1^n p_j\le n-(1-p)$ and hence, by the AM–GM inequality,
\begin{equation}
d\ge p-\Bigl(\frac{n-(1-p)}n\Bigr)^n=p-\Bigl(1-\frac{1-p}n\Bigr)^n=:f(p).
\end{equation}
Since $f(p)$ is increasing in $p\in[0,1]$, we have $f(p)\ge f(0)=-\bigl(1-\frac1n\bigr)^n$. So,
$d\ge-\bigl(1-\frac1n\bigr)^n$. On the other hand, if the sets $A_1^c,\dotsc,A_n^c$ form a partition of $\Omega$ and $P(A_j^c)=\frac1n$ for all $j$, then $p=0$ and $p_j=1-\frac1n$ for all $j$, whence $d=-\bigl(1-\frac1n\bigr)^n$.
So,
\begin{equation}
\min d=-\Bigl(1-\frac1n\Bigr)^n
\end{equation}
(so that $\min d\to-1/e$ as $n\to\infty$).