The answer given by David Speyer can be strengthened as follows. If $A$ is a non-invertible $n\times n$ matrix with entries in $R$ as described in the problem, then the linear maps $R^n \to R^n$ defined by either left or right multiplication are non-injective. In particular, $A$ is both a left-zero-divisor and a right-zero-divisor.
This is a consequence of McCoy's rank theorem. You can find a nice, brief account of this in Section 2 of this paper by Kodiyalam, Lam, and Swan. One consequence of the theorem is that for any commutative ring $R$, a square matrix $A$ over $R$ has linearly independent columns if and only if its determinant is not a zero-divisor, if and only if its rows are linearly independent.
So if every element of $R$ is either invertible or a zero-divisor, this means that every square matrix over $R$ defines a linear transformation that is either invertible or non-injective.
$\newcommand{\trace}{\operatorname{trace}}$
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 $s^2 \le nM^2 - m^2$, 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^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 $m^2$).
Best Answer
Bound 1 does not hold already for $n=2$. Take a matrix $A=\pmatrix{e^{ia}&e^{ib}\\e^{ic}&e^{id}}$, it satisfies your conditions if $|a+d-b-c|\leqslant \pi/3$. On the other hand, $X=\pmatrix{\cos a&\cos b\\\cos c&\cos d}$ and $$\det X=\cos a\cos d-\cos b\cos c=\frac12\left(\cos(a+d)+\cos(a-d)-\cos(b+c)-\cos(b-c)\right)\\= \frac12\left(2\sin\frac{b+c-a-d}2\sin\frac{a+d+b+c}2+\cos(a-d)-\cos(b-c)\right), $$ thus if $a-d=0$, $b-c=\pi$, $a+d+b+c=\pi$, $b+c-a-d=\pi/3$ (I am lazy to solve this explicitly), this expression is equal to $3/2>1$.