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.