Let's define the minimal linearly dependent subset of a matrix as below
Definition: For a matrix $A_{m\times n}=(\boldsymbol{c}_1, \ldots, \boldsymbol{c}_n)$ in which $\boldsymbol{c}_i$s are $m$-dimensional vectors, each set $\mathcal{S}=\{\boldsymbol{c}_{k_1}, \ldots, \boldsymbol{c}_{k_t}\}$ for a sequence $k_1, \ldots, k_t\in [0:n]$ is called a $t$-dimensional minimal linearly dependent subset of $A$, if vectors in $\mathcal{S}$ are linearly dependent but every $t-1$ members of $\mathcal{S}$ are linearly independent.
As an example, in the matrix \begin{align}A=\left( \begin{array}{c c c c} a&0&0&b&0&0&0\\ 0&a&0&0&b&0&0\\ 0&0&a&0&0&b&0\\ 0&0&0&a&0&0&b \end{array}\right) \end{align} we have $\mathcal{S}_1=\{\boldsymbol{c}_1, \boldsymbol{c}_4, \boldsymbol{c}_7\}$, $\mathcal{S}_2=\{\boldsymbol{c}_2, \boldsymbol{c}_4\}$, and $\mathcal{S}_3=\{\boldsymbol{c}_3, \boldsymbol{c}_6\}$ as $3$- and $2$-dimensional minimal subsets of $A$.
Upon my observation, for banded Toeplitz matrices, there should be some strong proportionality between the minimum dimension and the maximum number of total minimal linearly dependent subsets.
For example, as we see above, if we consider an $m\times (m+3)$ matrix formed by shifting $(a, 0, 0, b)$, there exists only 3 different subsets, each with dimension at least $\frac{m+2}{3}$. I guess there should be proportionality in the sense that, if we consider all types of banded Toeplitz, matrices, we have an increased number of higher-dimensional minimal subsets. What I like to prove is that the number is not that much, while the dimension grows.
Any idea on how can we find a tight relationship?