Just a brief addition to Denis's answer:

The function that you have is convex for unitarily invariant norms, but for the (basis dependent) elementwise absolute value, it can clearly break.

Here is a simple counterexample:

\begin{equation*} X = \begin{pmatrix} 2 & 0 & 0\\\\ 0 & 1 & 1\\\\ 0 & 0 & 1 \end{pmatrix},\qquad Y = \begin{pmatrix} 10 & 9 & 5\\\\ 9 & 10 & 5\\\\ 5 & 5 & 4 \end{pmatrix} \end{equation*}

Now, simply define $Z = 0.5(X+Y)$. Then, we see that $f(Z) = 33$, while $0.5(f(X)+f(Y)) = 3$, clearly violating convexity.

The function that you have is convex for unitarily invariant norms, but for the (basis dependent) elementwise absolute value, it can clearly break as a trivial counterexample below shows.

\begin{equation*} X = \begin{pmatrix} 2 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & 1 \end{pmatrix},\qquad Y = \begin{pmatrix} 10 & 9 & 5\\\\ 9 & 10 & 5\\\\ 5 & 5 & 4 \end{pmatrix} \end{equation*}

Now, simply define $Z = 0.5(X+Y)$, and consider $g(H)=\|H^{-1}\|_1$ as your function. Then, we see that $g(Z) = 3.1692$, while $0.5(g(X)+g(Y)) = 3$, clearly violating convexity.