Here is finally a solution for the $n=7$ case! Putting $T=2$, we have $c(M)=0$, where
$$M=\pmatrix
{1&T^{17}&T^{{32}}&T^{45}&T^{56}&T^{65}&T^{72}
\\0&1&\color{red}2\cdot T^{17}& T^{32}&T^{45}&T^{56} &T^{65}
\\0&0&1&\color{red}{\frac98}\cdot T^{17}& \color{red}{\frac{171}{64}} \cdot T^{32}&T^{45}&T^{56}
\\0&0&0&1&\color{red}{\frac98}\cdot T^{17} &T^{32}&T^{45}
\\0&0&0&0&1&\color{red}2\cdot T^{17}&T^{{32}}
\\0&0&0&0&0&1&T^{17}
\\0&0&0&0&0&0&1}
$$

Similarly to what I had said in the comments of S. Carnahan's answer, this is again based on a triangular matrix $A$ with constant diagonals, given by $A_{ij}=T^{k(18-k)}$ where $k=j-i$ for $j\geqslant i$. **For this kind of matrices, the TS seems to have *very* few complex roots (why?)**, e.g. $m(A)=42$ which is quite a bit less than $m(I_7)=8796$, and with the above adjustments the complex roots of $A$ can be eliminated.  
Note that like in the solution for $n=6$ given by S. Carnahan, the first diagonal is not concave.