Trace and determinant bounded by the operator norm

Trace. In the standard basis, $\operatorname{tr}A=\sum_{i=1}^n e_i^TAe_i$. For each $i$, by Cauchy-Schwarz,
$$|e_i^TAe_i|\le|e_i|\,|Ae_i|\le\|A\|\,|e_i|^2=\|A\|.$$
Summing the $n$ terms gives $|\operatorname{tr}A|\le n\|A\|$.

Determinant. Let $A=U\Sigma V^T$ be a singular value decomposition, with $U,V$ orthogonal and $\Sigma=\operatorname{diag}(\sigma_1\ge\dots\ge\sigma_n\ge0)$. Orthogonal matrices have determinant $\pm1$, hence
$$|\det A|=|\det U|\,\det\Sigma\,|\det V|=\sigma_1\sigma_2\cdots\sigma_n\le\sigma_1^n=\|A\|^n.$$

Check. For $A=\begin{pmatrix}2&0\\0&1\end{pmatrix}$: $\|A\|=2$, $\operatorname{tr}A=3\le4$ and $\det A=2\le4$.

Remark. Both bounds are attained by $A=I$ (trace) and by any orthogonal matrix times a scalar (determinant). The inequality $|\det A|\le\|A\|^n$ says that $\det$ is a polynomial map of degree $n$ with a controlled growth, which is useful to prove its continuity and to estimate the distance of a matrix to the singular matrices.