Assuming that the Hermitian minor has distinct eigenvalues, there is nothing we can say about the eigenvalues of $A$. Let the eigenvalues of the Hermitian minor be $\lambda_1$, ..., $\lambda_{n-1}$, so we can choose a matrix where your matrix looks like
$$A=\begin{pmatrix}
\lambda_1 & & & & a_1 \\
& \lambda_2 & & & a_2 \\
& & \lambda_3 & & a_3 \\
& & & \ddots & & \\
b_1 & b_2 & b_3 & & c
\end{pmatrix}$$

The characteristic polynomial of $A$ is
$$f(x):=(x-c) \prod_{i} (x-\lambda_i) - \sum_j a_j b_j \prod_{i \neq j} (x-\lambda_i). \quad (\ast)$$
I claim that we can choose $c$ and $a_j b_j$ to make $f(x)$ be any monic degree $n$ polynomial.

**Proof:** We have $f(\lambda_j) = - a_j b_j \prod_{i \neq j} (\lambda_i - \lambda_j)$. So, assuming that the $\lambda$'s are distinct, we can choose $a_j b_j$ to make $f(\lambda_j)$ have any value. Also, we can use $c$ to fix the value of $f$ at any $x$ other than the $\lambda_i$. A monic polynomial of degree $n$ is determined by its values at $n$ points, so we can arrange for $f$ to be any degree $n$ monic polynomial. $\square$

Equation $(\ast)$ also shows that, if $\lambda_i$ occurs with multiplicity $d$ in the Hermitian minor, then it occurs with multiplicity $\geq d-1$ in $A$.