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> Is it possible to prove this matrix family only contains totally unimodular matrices?
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The matrix has dimensions $\frac{3n(n-1)}2$ rows and $n+\frac{n(n-1)}2$ columns.


To every pair $(i,i')$ with $1\leq i<i'\leq n$ we have an unique integer $f(i,j)$ from $\big\{n+1,\dots,n+\frac{n(n-1)}2\big\}$ associated with it.


1. Each $3r+1$ row has three non-zero entries with $M\_{(3r+1),i}=M\_{(3r+1),i'}=1$ at some $1\leq i<i'\leq n$ (two of first $n$ columns are $1$) and $M\_{(3r+1),f(i,i')}=1$.
2. Each $3r+2$ and $3r+3$ row has two non-zero entries $M\_{(3r+2),i}=M\_{(3r+2),f(i,i')}=M\_{(3r+3),i'}=M\_{(3r+3),f(i,i')}=1$.


$$M=\begin{bmatrix}1&1&0&1&0&0\\1&0&0&1&0&0\\0&1&0&1&0&0\\1&0&1&0&1&0\\1&0&0&0&1&0\\0&0&1&0&1&0\\0&1&1&0&0&1\\0&1&0&0&0&1\\0&0&1&0&0&1\end{bmatrix}$$ holds at $n=3$ and it is totally unimodular here.



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> Is this type of matrix always totally unimodular?
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One can prove by induction if we know that if $A,B$ are totally unimodular then $\begin{bmatrix}A\\B\end{bmatrix}$ is totally unimodular under mild non-intersection conditions.



Is this totally unimodular family?
I don't see that your example matrix $M$ is TU. Taking the first three columns of rows 1, 4 and 7 gives the submatrix
$$\begin{pmatrix}1&1&0\\ 1&0&1\\ 0&1&1\end{pmatrix}$$
which has derterminant $-2$.


