An upper bound of the determinant

Idea. Hadamard bounds the determinant by the product of column lengths; AM-GM then bounds this product by the arithmetic mean of the squared lengths, which is $\|A\|_F^2/n$.

Step 1: Hadamard. $|\det A|^2\le\prod_{j=1}^n|a_j|^2$.

Step 2: AM-GM. For the non-negative numbers $|a_j|^2$,
$$\prod_{j=1}^n|a_j|^2\le\Big(\frac1n\sum_{j=1}^n|a_j|^2\Big)^n=\Big(\frac{\|A\|_F^2}n\Big)^n,$$
since $\sum_j|a_j|^2=\sum_{i,j}a_{ij}^2=\|A\|_F^2$. Taking square roots of the combined inequality gives the claim.

Equality, if $\det A\ne0$. We need equality in both steps. Equality in Hadamard means the columns are pairwise orthogonal; equality in AM-GM means all $|a_j|^2$ are equal, say to $s^2>0$. Then $A^TA=s^2I$, so $A/s$ is orthogonal and $A=sU$ with $U\in O(n)$.

Equality, if $\det A=0$. The left side is $0$, so the right side must vanish: $\|A\|_F=0$, i.e. $A=0$.

Converse. If $A=sU$ then $|\det A|=|s|^n$ and $\|A\|_F=|s|\sqrt n$, so both sides equal $|s|^n$.

Remark. Writing the inequality as $|\det A|^{2/n}\le\frac1n\|A\|_F^2=\frac1n\operatorname{tr}(A^TA)$ shows it is the AM-GM inequality for the eigenvalues of the PSD matrix $A^TA$ (whose determinant is $\det(A)^2$ and whose trace is $\|A\|_F^2$): $\det(S)^{1/n}\le\operatorname{tr}(S)/n$.