The smallest nonzero eigenvalue can decrease at least exponentially,
even for matrices that are sparse, symmetric, and invertible.

Explicitly, let $M_n$ have $1$'s on the *anti*-diagonal, and also on the
first and third off-diagonals above it. For example,
here's the matrix for $n=13$:

```
0 0 0 0 0 0 0 0 0 1 0 1 1
0 0 0 0 0 0 0 0 1 0 1 1 0
0 0 0 0 0 0 0 1 0 1 1 0 0
0 0 0 0 0 0 1 0 1 1 0 0 0
0 0 0 0 0 1 0 1 1 0 0 0 0
0 0 0 0 1 0 1 1 0 0 0 0 0
0 0 0 1 0 1 1 0 0 0 0 0 0
0 0 1 0 1 1 0 0 0 0 0 0 0
0 1 0 1 1 0 0 0 0 0 0 0 0
1 0 1 1 0 0 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0
```

Then (much as in
this answer)
the inverse matrix $M_n^{-1}$ is anti-triangular with constant antidiagonals;
thus it is determined by its bottom row, and this bottom row is
$1, -1, 1, -2, 3, -4, 6, -9, 13, -19, 28, -41, \ldots$,
with alternating signs and absolute values satisfying the recurrence
$t_m = t_{m-1} + t_{m-3}$. Thus $t_m$ grows like a multiple of $C^m$
where $C = 1.46557\ldots$ is the real root of $C^3 = C^2 + 1$,
and the main diagonal of $M_n^{-1}$ has constant sign. Here is
$M_{13}^{-1}$:

```
0 0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 1 -1
0 0 0 0 0 0 0 0 0 0 1 -1 1
0 0 0 0 0 0 0 0 0 1 -1 1 -2
0 0 0 0 0 0 0 0 1 -1 1 -2 3
0 0 0 0 0 0 0 1 -1 1 -2 3 -4
0 0 0 0 0 0 1 -1 1 -2 3 -4 6
0 0 0 0 0 1 -1 1 -2 3 -4 6 -9
0 0 0 0 1 -1 1 -2 3 -4 6 -9 13
0 0 0 1 -1 1 -2 3 -4 6 -9 13 -19
0 0 1 -1 1 -2 3 -4 6 -9 13 -19 28
0 1 -1 1 -2 3 -4 6 -9 13 -19 28 -41
1 -1 1 -2 3 -4 6 -9 13 -19 28 -41 60
```

Hence the trace of $M_n^{-1}$ grows as $\pm C^n$,
so its largest eigenvalue grows at least as $\pm C^n/n$.
Therefore the smallest eigenvalue of $M_n$ is $O(n/C^n)$.

(Numerical computation suggests that in fact
there's only one really small eigenvalue,
which is thus $O(C^{-n})$; for example, $M_{13}$ has an eigenvalue
$0.008902\ldots$, and the next-smallest eigenvalues are about
$-.78$ and $.82$.)

Here's some **gp** code to generate the matrix $M_n$
and the absolute value **m(n)** of its least eigenvalue:

```
M(n) = matrix(n,n,i,j, (i+j == n+1) + (i+j == n) + (i+j == n-2))
m(n) = vecmin(abs(real(polroots(charpoly(M(n))))))
```