The answer is no in general since Hermitian matrices have real eigenvalues. So for example, there is no Hermitian matrix whose characteristic polynomial is $X^2+1$.
Even if you restrict to polynomials with real roots, I doubt you can find a simple formula : I think that if there is a rational formula that works for every polynomial with real roots, it should also apply to all polynomials. Let me explain what I mean precisely. We'll use the following lemma which states that when rational equations hold for polynomial with real roots, they should hold everywhere. Denote $\mathbb{C}_{n,u}[X]$ the set of monic complex polynomials of degree $n$.
Lemma : Let $f$ be a rational function
$$\begin{eqnarray*}f :& \mathbb{C}_{n,u} & \longrightarrow \mathbb{C}^N\\
& P &\longmapsto f(P)\end{eqnarray*}$$
such that $f$ is zero on polynomial with real roots. Then $f$ is zero.
Proof :
The map
$$\begin{eqnarray*}p_n :& \mathbb{C}^n & \longrightarrow \mathbb{C}_{n,u}[X]\\
& (\lambda_1, \ldots, \lambda_n) &\longmapsto p_n(\lambda_1, \ldots, \lambda_n) = \prod_{i = 1}^n (X-\lambda_i)\end{eqnarray*}$$
is also a rational map (meaning the coefficients of a polynomial are rational functions of its roots, and we can actually compute them, those are called elementary symmetric polynomials). By hypothesis, the function $f \circ p_n$ is zero on $\mathbb{R}^n$. But since $f \circ p_n$ is rational, it is actually zero everywhere, and because $p_n$ is surjective (every polynomial is split in $\mathbb{C}$), then $f$ is zero.
Now assume there is a rational function
$$\begin{eqnarray*}A_n :& \mathbb{C}_{n,u}[X] & \longrightarrow \mathcal{M}_n(\mathbb{C})\\
& P &\longmapsto A_n(P)\end{eqnarray*}$$
(by that I mean that the coefficients of the matrix $A_n(P)$ are rational functions of the coefficients of $P$) such that the characteristic polynomial of $A_n(P)$ is $P$ whenever $P$ has real roots. This means the rational function $P \longmapsto \det(XI_n - A_n(P)) - P$ is zero on polynomial with real roots, so it's zero everywhere, which means the characteristic polynomial of $A_n(P)$ is always $P$ without assumption on the roots of $P$.
Assume moreover that $A_n(P)$ is hermitian whenever $P$ has real roots. Then I claim $A_n(P)$ is hermitian for any real polynomial $P$ from the same kind of argument. Indeed denote $A_n^* : P \longmapsto \left[A_n(P^*)\right]^*$, where $P^*$ denotes the complex conjugate of the polynomial $P$, and $\left[A_n(P^*)\right]^*$ is the conjugate transpose of the matrix $A_n(P^*)$. This is a rational function (beware that $P \mapsto [A_n(P)]^*$ is not rational however !). Our assumption is that the rational function $A_n - A_n^*$ is zero on all polynomial with real roots, so it's zero everywhere. When $P$ is real, this means $A_n(P)$ is hermitian (because $A_n(P) = A_n^*(P) = [A_n(P)]^*$, the last equality being only true when $P$ is real). This yields a contradiction if $P$ is a real polynomial with complex roots.
The determinant is a sum of (signature-weighted) products of $n$ elements, where no two elements share the same row or column index. From this, it follows, that there is no term with $(n-1)$ terms on the diagonal (if $n-1$ terms of a product are on the diagonal, then the last one must be too, because all other rows and columns are taken). So... the only term that can possibly include a power of $\lambda^{n-1}$ is the product of the main diagonal. Therefore, the $\lambda^{n-1}$ coefficient of $\det A$ equals the $\lambda^{n-1}$ coefficient of $\prod_i (\lambda-A_{ii})$ for which it's easy to show, the coefficient equals $-\sum_i A_{ii}$.
This definition doesn't make assumptions about the field over which the matrix is defined, because the field operations + and * are used directly (with no assumptions about inverses and distribution laws).
Best Answer
This is true. We have that $\frac{1}{i}=-i$, and if $X$ is skew-Hermitian, $-iX$ is Hermitian. To check this, denoting $X^H$ to be the conjugate transpose, we have $$ (-iX)^H=\overline{-i}(X^H)=i(-X)=-iX $$ Now, Hermitian matrices have all real eigenvalues, so the characteristic equation will also have all real coefficients.
Another way to see that all the eigenvalues of $-iX$ are real is to use the fact that skew-Hermitian matrices are diagonalizable, and move the $-i$ coefficient to the diagonal matrix, whose entries are the eigenvalues of $X$.