To just add to the answers already here, it is possible to extract the individual coefficients of this polynomial exactly using the following method. Let
\begin{equation}
h(x) = \sum_{j = 0}^d c_j x^j
\end{equation}
be a polynomial of degree $d$, then
\begin{equation}
c_j = \frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}})
\end{equation}
since
\begin{align}
\frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}}) &= \frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} \left(\sum_{l = 0}^d c_l e^{\frac{2 \pi i k l}{d + 1}} \right) \\
&= \sum_{l = 0}^d \left(\frac{1}{d + 1} \sum_{k = 0}^d e^{\frac{2 \pi i (l - j) k}{d + 1}} \right) c_l \\
&= \sum_{l = 0}^d \delta_{jl} c_l \\
\frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}}) &= c_j
\end{align}
where $\delta_{jl}$ is the Kronecker delta and $i$ is the imaginary unit. Letting
\begin{equation}
f(x, y, z) = \det(x A + y B + z C)
\end{equation}
we can expand $f$ in powers of $x, y, z$ using multilinearity of the rows or columns of the matrix inside the determinant as
\begin{equation}
f(x, y, z) = \sum_{j = 0}^n x^j g_j(y, z) = \sum_{j = 0}^n x^j \left(\sum_{k = 0}^{n - j} a_{j, k} y^k z^{n - j - k}\right)
\end{equation}
Note that, since one choice of $x, y, z$ must be made for every row or column in the expansion, the powers of these variables must sum to $n$ in every term. This polynomial is completely specified by the coefficients $\{a_{j, k}\}$, and to extract them, it is sufficient to only evaluate $f$ at $z = 1$
\begin{equation}
f(x, y, 1) = \sum_{j = 0}^n x^j g_j(y, 1) = \sum_{j = 0}^n \sum_{k = 0}^{n - j} a_{j, k} x^j y^k
\end{equation}
One query of the polynomial
\begin{equation}
g_j(y, 1) = \sum_{k = 0}^{n - j} a_{j, k} y^k
\end{equation}
is given by the above formula
\begin{equation}
g_j(y, 1) = \frac{1}{n + 1} \sum_{p = 0}^n e^{-\frac{2 \pi i j p}{n + 1}} f(e^{\frac{2 \pi i p}{n + 1}}, y, 1)
\end{equation}
Similarly,
\begin{equation}
a_{j,k} = \frac{1}{n - j + 1} \sum_{q = 0}^{n - j} e^{-\frac{2 \pi i k q}{n - j + 1}} g_j(e^{\frac{2 \pi i q}{n - j + 1}}, 1)
\end{equation}
Therefore
\begin{equation}
\det(x A + y B + z C) = \sum_{j = 0}^n \sum_{k = 0}^{n - j} a_{j,k} x^j y^k z^{n - j - k}
\end{equation}
where
\begin{equation}\
a_{j, k} = \frac{1}{(n + 1)(n - j + 1)}\sum_{p = 0}^{n} \sum_{q = 0}^{n - j} e^{-2 \pi i \left(\frac{jp}{n + 1} + \frac{kq}{n - j + 1} \right)} \det(e^{\frac{2 \pi i p}{n + 1}} A + e^{\frac{2 \pi i q}{n - j + 1}} B + C)
\end{equation}
requiring the evaluation of only $O(n^4)$ determinants to completely characterize the polynomial.
