Partial answer: As Christian Remling noted in remarks, it suffices to deal with the case that $(A-tI_{n})^{2} = \lambda I_{n}.$ Since $A$ is symmetric, we can only have $\lambda = 0$ when $A = tI_{n}$ for some $t$. But since the diagonal entries of $A$ are all $0,$ this only happens when $A = 0.$ Hence we may suppose that $\lambda \neq 0.$
 Then $A = tI_{n} + \sqrt{\lambda}T$ for some matrix $T$ with $T^{2} = I_{n}.$
If $T = I_{n}$ we again see that $A$ is a scalar matrix, so the zero matrix.
   Hence the only non-zero possibilities for $A$ arise when $T^{2} = I \neq T.$ 
Now trace $T$ = $r-s$ where $T$ has $r$ eigenvalues $1$ and $s$ eigenvalues $-1$
Also, since $tI_{n} + \sqrt{\lambda}T$ has all diagonal entries zero, it follows that all diagonal entries of $T$ must be equal. Hence each diagonal entry of $T$ is $\frac{r-s}{n}$ and $t + \frac{(r-s)\sqrt{\lambda}}{n} = 0.$ 

Now $r = s$ gives $t = 0,$ so that $A = \sqrt{\lambda}T.$  If $r \neq s,$ so that $\lambda = \frac{n^{2}t^{2}}{r-s}.$

Hence the problem is reduced in essence to determining which symmetric matrices $T$ of order two have all diagonal entries equal. But if $T$ is such a matrix,
then $\frac{I+T}{2}$ and $\frac{I-T}{2}$ are mutually orthogonal idempotent symmetric matrices, each with all diagonal entries equal.

Now the problem is reduced to finding symmetric idempotent matrices $E$ with all diagonal entries equal (for we may take $T = 2E-I$ if we find such an $E$). If $E$ has rank $m,$ then note that each diagonal entry of $E$ is $\frac{m}{n}.$

We may obtain such an $E$ or each positive divisor $d$ of $n$: take $E$ to be the direct sum (in the obvious sense) of $n/d$ copies of $\frac{J_{d}}{d},$ where $J_{d}$ is the $d \times d$ matrix with all entries $1.$

At the moment, I don't see how to determine all possibilities for such an $E$.