Say the matrix $S$ has unitary orthonormal eigenvectors $u_1,u_2,\dots u_n$ with eigenvalues
$$\lambda_1\leq \lambda_2\leq \dots \leq \lambda_{n-1}\leq \lambda_n$$
I think the smallest (resp. largest) eigenvalue of a principal minor is $\leq \lambda_2$ (resp. strictly $\geq \lambda_{n-1}$).

A sufficient condition for this is to have ortho-normal eigenvectors in $S$ and in its minors (true in a positive definite
world with symmetric matrices).

For any chosen coordinate $i$, the idea is to consider a linear combination of $v=au_1+bu_2$ such that $v_i=0$ as suggested above
(if I understood well the comment of T. Tao). I think one can prove that $\frac{|Sv|}{|v|}\in [\lambda_1,\lambda_2]$,
where $|v|^2=a^2+b^2$ (to verify it, write $Sv=a\lambda_1 u_1+b\lambda_2 u2$ and observe $|Sv|^2= \lambda_1^2a^2 +
\lambda_2^2b^2 $. In other words, the multiplication by $S$ ''stretches'' $v$ by something between $\lambda_1$ and $\lambda_2$.
Now, we can see how the minor $S^i$ ($S$ without line and column $i$) acts on $v^i$ ($v$ without position $i$). Column $i$ does nothing to $v$ in the calculation of $Sv$, and so, $S^iv^i$ is the same as vector $Sv$ but without
the (non-zero) value of $Sv$ at coordinate $i$. By forgetting this coordinate, $S^iv^i$ has an even smaller 2-norm than $Sv$. Vector
$v^i$ is ``stretched'' via the $S^i$ multiplication by something $\leq \lambda_2$. As such, $S^i$ can not have all eigenvalues larger than $\lambda_2$, this would mean that it
''stretches'' any $(n-1)-$dimensional vector (including $v^i$) by more than $\lambda_2$.

Finally, the inequalities are not strict because $a$ or $b$ above can be zero. The very basic matrix
$\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}$
has eigenvalues 0 and 1 but the right-left minor has eigenvalue 1.