The trace is bounded by the Frobenius norm

Idea. The trace only sees the $n$ diagonal entries, while the Frobenius norm sees all $n^2$ entries; Cauchy-Schwarz compares the sum of $n$ numbers with the length of the vector they form.

Step 1: Cauchy-Schwarz. Using $1\cdot a_{ii}$ in $\mathbb{R}^n$,
$$|\operatorname{tr}A|=\Big|\sum_{i=1}^n1\cdot a_{ii}\Big|\le\Big(\sum_i1^2\Big)^{1/2}\Big(\sum_ia_{ii}^2\Big)^{1/2}=\sqrt n\Big(\sum_ia_{ii}^2\Big)^{1/2}.$$

Step 2: add the off-diagonal terms. Since $\sum_ia_{ii}^2\le\sum_{i,j}a_{ij}^2=\|A\|_F^2$, we get $|\operatorname{tr}A|\le\sqrt n\|A\|_F$.

Equality. Both steps must be equalities: the off-diagonal entries vanish and the diagonal vector $(a_{ii})$ is proportional to $(1,\dots,1)$. So equality holds iff $A=cI$ is a scalar matrix.

Second proof. $\operatorname{tr}A=\langle I,A\rangle_F$, and by Cauchy-Schwarz in $(M_n,\langle,\rangle_F)$, $|\langle I,A\rangle_F|\le\|I\|_F\|A\|_F=\sqrt n\|A\|_F$.

Remark. The trace is a linear functional, hence continuous, and this computation gives its norm: $\|\operatorname{tr}\|=\sqrt n$ with respect to $\|\cdot\|_F$ (attained at $A=I$). With respect to the operator norm, $|\operatorname{tr}A|\le n\|A\|$ (also attained at $I$).