Trace norm of a triangular matrix with only ones above the diagonal

In Loewner ordering, we have $$ T_n^{-1}(T_n^{-1})^\top =\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&1} \preceq\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&2}=P. $$ Using the spectral formula for tridiagonal Toeplitz matrices, the eigenvalues of $P$ are given by $2+2\cos\left(\frac{k\pi}{n+1}\right)=4\cos^2\left(\frac{k\pi}{2(n+1)}\right)$ with $k=1,2,\ldots,n$. Therefore, the $k$-th smallest singular value of $T_n$ is bounded below by $\frac12\sec\left(\frac{k\pi}{2(n+1)}\right)$ and \begin{align} \frac1n\|T_n\|_{\text{Tr}} &\ge\frac1n\sum_{k=1}^n\frac12\sec\left(\frac{k\pi}{2(n+1)}\right)\\ &\ge\frac1\pi\int_0^{n\pi/2(n+1)}\sec x\,dx\\ &=\frac1\pi\ln\left(\sec\frac{n\pi}{2(n+1)} + \tan \frac{n\pi}{2(n+1)}\right), \end{align} which is unbounded when $n\to\infty$.