Let $A$ be the linear map on the space of complex sequences acting as $$(Au)_{n}=u_{n-1}+a_{n}u_{n}+u_{n+1}, \quad n\in\mathbb{Z},$$ where $\{a_{n}\}$ is a fixed sequence. Let $f=f(z)$ and $g=g(z)$ be two solutions of the eigenvalue equation $Au=zu$ for all $z\in\mathbb{C}$. Suppose also that $f_n(z)$ and $g_n(z)$, as functions of $z$, have all derivatives w.r.t. $z$ (denoted by an upper index or a prime in the following text).
One easily verifies that (discrete Wronskian or Casoratonian) $$C_{n}(f,g):=f_{n+1}g_{n}-f_{n}g_{n+1}$$ does not depend on $n$. Thus, I will drop the index and write only $C(f,g)$.
It seems the determinant of the matrix $$ M_{k,n}=\begin{pmatrix} f_{n} & f'_{n} & \dots & f^{(k)}_{n} & g^{(k)}_{n}\\ f_{n+1} & f'_{n+1} & \dots & f^{(k)}_{n+1} & g^{(k)}_{n+1}\\ \vdots & \vdots & & \vdots & \vdots\\ f_{n+k+1} & f'_{n+k+1} & \dots & f^{(k)}_{n+k+1} &g^{(k)}_{n+k+1}\\ \end{pmatrix} $$ is expressible in terms of $C(f,g)$ and its derivatives up to order $k$, perhaps in the form $$ \det M_{k,n}=\sum_{j=0}^{k}\alpha_{j,n}(f,f',\dots,f^{(k)})C^{(j)}(f,g). $$
Can you see/deduce the general formula? Is it well-known?