The question concerns the determinant of a Hankel matrix, or a fixed element of a Hankel matrix transform of a shifted sequence $a(n,k)$ for a fixed $k$, although I do not see how this fact alone can be useful. I give a standalone proof below.

Let $k$ be fixed. For the sake of simplicity, let's denote $c_j := c(k,j)$ and $a_n := a(n,k)$.

First, notice that the sequence $(a_n)$ has the characteristic polynomial
$$C(x) := \sum_{j=0}^k c_j x^{k-j},$$
while $C(-x)$ gives the characteristic polynomial for the sequence $((-1)^n a_n)$.

Now, let $b_t(n)$ for $t\in\{0,\dots,k-1\}$ be the determinant of the matrix obtained from the $(k-1)\times k$ matrix: $$\big(a_{n+i+j-k}\big)_{1\leq i\leq k-1\atop 0\leq j\leq k-1}$$
by removing the column with $j=t$.
Then $\det A_k(n) = b_0(n) = b_{k-1}(n+1)$.

From the recurrence for $a_n$, it follows that
$$
\begin{bmatrix} b_0(n) \\ b_1(n)\\ \vdots\\ b_{k-1}(n) \end{bmatrix} = M\cdot
\begin{bmatrix} b_0(n-1) \\ b_1(n-1)\\ \vdots\\ b_{k-1}(n-1) \end{bmatrix},$$
where
$$
M := \begin{bmatrix}
(-1)^{k-1} c_{k-1} & (-1)^{k-1} c_k & 0 & \dots & 0\\
(-1)^{k-2} c_{k-2} & 0 & (-1)^{k-1} c_k & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-c_1 & 0 & 0 & \dots & (-1)^{k-1} c_k\\
1 & 0 & 0 & \dots & 0
\end{bmatrix}.
$$
This is almost a companion matrix and its characteristic polynomial equals the reciprocal of $C((-1)^k x)$, which is same as $C((-1)^k x)$ thanks to the symmetry. In particular, $b_0(n)$ has a characteristic polynomial $C(x)$ when $k$ is even, $C(-x)$ when $k$ is odd.

It remains to notice that $b_0(n)=(-1)^{(k-1)(k-2)/2} a_n$ for $n=-(k-1),\dots, 0$, implying that $\det A_k(n) = b_0(n) = (-1)^{kn + (k-1)(k-2)/2} a_n$. QED

**UPDATE.** More generally, we can drop the symmetricity requirement $c_j = c_{k-j}$ and keep only $c_0 = c_k$. Then
$$\det A_k(n) = (-1)^{kn + (k-1)(k-2)/2} a_{-n-k},$$
where $a_n$ is extended to large negative indices by the same recurrence (see also this answer). This formula implies the original result, since for symmetric $c_j$, we have $a_{-n-k}=a_n$.

Similarly, if we have $c_0 = -c_k$, then
$$\det A_k(n) = (-1)^{(k+1)n + (k+1)k/2} a_{-n-k}.$$