Consider a unitary matrix $U = (u_{i,j})$. We will show the slightly stronger statement that there is some permutation $\pi:[n] \to [n]$ such that $|u_{i,\pi(i)}| \leq 1/\sqrt{n}$ for all $i$. Note that the process of permuting columns doesn't preserve the trace of the matrix, but it does preserve the property of being unitary!

Given $\pi$, consider the associated quantity $I(\pi) = \sum_{i=1}^n |u_{i,\pi(i)}|^2$. Since there are only finitely many permutations, there is some permutation $\pi_*$ minimizing this value. In particular, letting $v_{i,j} = u_{i,\pi_*(j)}$, the resulting matrix $V= (v_{i,j})$ after applying the permutation of the columns must satisfy
$$|v_{i,j}|^2 + |v_{j,i}|^2 \geq |v_{i,i}|^2 + |v_{j,j}|^2$$
for all indices $i,j$, since otherwise we could exchange rows $i$ and $j$ and obtain a smaller value of $I(\pi)$.

Consider summing the above inequality over $j$. On the left hand side we get the squared norm of a row and a column of $V$, which both must be $1$. Thus we get:
$$2 \geq n|v_{i,i}|^2 + I(\pi_*).$$
Summing over $i$, we moreover deduce that:
$$2n \geq 2nI(\pi_*),$$
whence we conclude that $I(\pi_*) \leq 1$ and hence from the previous inequality $n |v_{i,i}|^2 \leq 1$. Rearranging gives the desired result.