$\newcommand{\trace}{\operatorname{trace}}$ The result below mentions a reasonably improved inequality.
Let $m = \frac{\trace(A)}{n}$, and $s^2= \frac{\trace(A^2)}{n}-m^2$. Then, Wolkowicz and Styan (Linear Algebra and its Applications, 29:471-508, 1980), show that
\begin{equation*} \lambda_1 \ge \frac{\det(A)}{(m+s/\sqrt{n-1})^{n-1}} \end{equation*}
Remark: As per the notation in the OP, $\lambda_1$ is the smallest eigenvalue---usually the literature uses $\lambda_1$ to be largest.
Thus, we obtain the upper bound \begin{equation*} \lambda_2\lambda_3\cdots\lambda_n \le \left(m+ \frac{s}{\sqrt{n-1}}\right)^{n-1}. \end{equation*} This bound is tight. Consider for example, If $A= \text{Diag}(1,2,2,\ldots,2)$, then the lhs is $2^{n-1}$, $m=2-1/n$, and $s^2 = 1/n-1/n^2$, so that $s/\sqrt{n-1} = 1/n$. Thus, the bound on the rhs is tight.
With $M := \max_{i,j}|a_{ij}|$, we see that $m \le M$ and $s^2 \le nM - m^2$$s^2 \le nM^2 - m^2$, which leads to an upper bound in terms of $M$ as desired
\begin{equation*} \lambda_2\lambda_3\cdots\lambda_n \le \left(M+ \sqrt{\frac{nM-m^2}{n-1}}\right)^{n-1} < \left(M + \sqrt{\frac{nM}{n-1}} \right)^{n-1}, \end{equation*}\begin{equation*} \lambda_2\lambda_3\cdots\lambda_n \le \left(M+ \sqrt{\frac{nM^2-m^2}{n-1}}\right)^{n-1} < \left(M + M\sqrt{\frac{n}{n-1}} \right)^{n-1}, \end{equation*} which is better than the bound mentioned in the post (though we lost a bit by deleting $m^2$).