Potential p-norm on tuples of operators

No, the expression $$ left|left[begin{matrix}A \ Bend{matrix}right]right|_p = | |A|^p + |B|^p|^{1/p}. $$ is not a norm for any $2<p<infty$ (so it is a norm if and only if $p=2,infty$).

I will justify that by proving that the triangle inequality for this expression would imply that the map $tmapsto t^{p/2}$ is operator monotone, which is well-known to be false (Loewner).

Lemma. Given $A_1,A_2$ bounded positive operators on a Hilbert space, the following are equivalent:

1. $|A_1+B| leq |A_2+B|$ for every positive operator $B$.
2. $A_1 leq A_2$.

Proof: one direction ($2. implies 1.$) is obvious. For the other, let $u$ be a unit vector, $P_u$ the rank one orthogonal projection on $mathbf{C}u$ and take $B= s P_u$ for $s>0$ (large). Then a small computation gives that $|A_1+B| = s + langle A_1 u,urangle + O(1/s)$, so making $s to +infty$, we obtain that 2. implies that $langle A_1 u,urangle leq langle A_2 u,urangle$. The lemma is proven.

Assume for a contradiction that your expression was a norm. Then for any $C$ of norm $<1$, we can write $C$ as the average of two unitaries (see here), and therefore (for arbitrary $A,B$) we can write $left[begin{matrix} CA \ Bend{matrix}right]$ as a convex combination of elements of the form $left[begin{matrix}UA \ Bend{matrix}right]$ for unitares $U$, which all have the same "norm" $| |A|^p + |B|^p|^{1/p}$, and therefore we would have $$| |CA|^p+|B|^p| ^{1/p}leq | |A|^p + |B|^p|^{1/p}.$$ So by the Lemma we would have $$ |CA|^p leq |A|^p$$ for any $A$ and any $C$ of norm $leq 1$. Note that every operator $leq |A|^2=A^* A$ can be written as $|CA|^2$ for some $C$ of norm $leq 1$. So we have reached the desired conclusion that the map $tmapsto t^{p/2}$ is operator monotone.