Every real symmetric matrix is similar to a diagonal matrix by the spectral theorem. The only nilpotent matrix which
is similar to a diagonal matrix is the zero matrix. Hence if you mean principal submatrices then the answer is
trivial, as Chris Godsil has pointed out.

If you consider arbitrary submatrices, the question seems to be quite vague.
For finding in the sense of *counting* nilpotent symmetric matrices, it seems to be more interesting to consider
other fields than $\mathbb{R}$. Indeed, let the field be $\mathbb{F}_q$, and let $a(n,k,q)$ be the number of
nilpotent matrices of rank $k$ over $\mathbb{F}_q$. Then
$$
a(n,k,q)=\frac{\Phi_n(q)\Phi_{n-1}(q)}{\\Phi_{n-k}(q)\Phi_{n-k-1}(q)\Phi_k(q)}q^{\binom{k}{2}}
$$
for $k \le n-1$, where $\Phi_r(q)=\prod_{1\le i\le r}(q^i-1)$. This is due to G. Lusztig. Then count the nilpotent,
symmetric ones of rank $k$, see http://www.win.tue.nl/~aeb/math/symnilp.html .