The claim is true. We prove it using a few block matrix manipulations. Note, in the proofs below $A \ge 0$ means $A$ is (symmetric) positive semidefinite.
$\newcommand{\trace}{\text{trace}}$
Lemma Let $X, Y \ge 0$. Then,
\begin{equation*}
\otimes^k (X+Y) \ge \otimes^k X + \otimes^k Y.
\end{equation*}
Proof.
By induction on $k$. The case $k=1$ is trivial; $k=2$ shows us the crux. Indeed,
\begin{equation*}
(X+Y)\otimes (X+Y) - X\otimes X - Y\otimes Y = X\otimes Y + Y \otimes X \ge 0,
\end{equation*}
since $X, Y \ge 0$. The general case follows similarly.
-
Corollary.
Let $X, Y \ge 0$. Then, by restricting to the suitable symmetry class of tensors we get
\begin{equation*}
\wedge^k(X+Y) \ge \wedge^k X + \wedge^k Y
\end{equation*}
.
Lemma. Let $A=[A_{ij}]$ be $mn\times mn$ with $n\times n$ blocks. Suppose $A$ is symmetric, positive semidefinite. Then, for $1\le k \le n$, the $m\binom{n}{k} \times m\binom{n}{k}$ matrix $C_k := [\wedge^k A_{ij}]$ is semidefinite.
Proof. Some reflection shows that $C_k$ is a principal submatrix of $\wedge^k A$, thus, $C_k = P_k^*(\wedge^k A)P_k \ge 0$ since $A\ge 0$ and wedge products preserve positivity.
Theorem. Let $A=[A_{ij}] \ge 0$ and $B=[B_{ij}] \ge 0$ be $mn\times mn$ block matrices composed of $n\times n$ blocks. Define $$M_k = [\trace(\wedge^k(A_{ij}+B_{ij}))] - [\trace(\wedge^k A_{ij})] - [\trace(\wedge^k B_{ij})],$$ for any $1\le k \le n$. Then, $M_k \ge 0$.
Proof
The Corollary above shows that $\wedge^k(A+B) \ge \wedge^k A + \wedge^k B$. Let $P_k$ be as in the second Lemma; then $$H_k = P_k^*(\wedge^k(A+B))P_k- P_k^*(\wedge^k A)P_k -P_k^*(\wedge^k B)P_k \ge 0.$$ The matrix $M_k$ is nothing but a (blockwise) partial trace of $H_k$, so that $H_k \ge 0 \implies M_k \ge 0$.
Corollary
Let $A$ and $B$ be as above. Then,
\begin{equation*}
M = [\det(A_{ij}+B_{ij})] - [\det A_{ij}] - [\det B_{ij}] \ge 0.
\end{equation*}
Proof.
Observe that $\trace(\wedge^n X) = \det(X)$ for an $n\times n$ matrix $X$.
