By the standard classification of [unitarily invariant norms (see e.g., this blog post][1], the expression $N_{(p,q)}$ is a norm if and only if the function
$$ \| (x,y) \| := \inf \left\{ \varepsilon > 0: \left(\frac{\max(|x|, |y|)}{\varepsilon}\right)^p + \left(\frac{\min(|x|, |y|)}{\varepsilon}\right)^q \leq 1 \right\}$$
is a norm, or equivalently if the set
$$ \{ (x,y): \max(|x|,|y|)^p + \max(|x|,|y|)^q \leq 1 \}$$
is convex (certainly it is symmetric and has non-empty interior).  It's slightly more convenient to rescale this to
$$ S := \{ (x,y): \max(|x|,|y|)^p + \max(|x|,|y|)^q \leq 2 \}.$$
In the upward sector $\{ (x,y): |x| \leq |y| \}$, this set is convex from the convexity of $a^p+b^q$ already observed by the OP, and at the endpoint $(1,1)$ of the set on the upper right edge of the sector, implicit differentiation of the condition $y^p + x^q = 2$ reveals that the boundary of this set has a slope of $-\frac{q}{p}$ at this point, which is less than or equal to $-1$ iff $p \geq q$.  Similarly at the opposite endpoint $(-1,1)$, the slope is at most $+1$ iff $p \geq q$.  Hence the entire set $S$ is convex iff $p \geq q$, and so the original expression $N_{(p,q)}$ is a matrix norm iff $p \geq q$.

For instance, in the limit as $p \to \infty$ this norm converges to the operator norm (regardless of what $q$ is doing in this limit).

To obtain an explicit counterexample to the triangle inequality when $p < q$: observe that $\mathrm{diag}(1,1)$ is in the convex hull of the matrices $\mathrm{diag}(1+\delta,1-\delta)$, $\mathrm{diag}(1-\delta,1+\delta)$ for any $0 < \delta < 1$, so by symmetry and the triangle inequality one needs
$$ N_{(p,q)}(\mathrm{diag}(1,1)) \leq N_{(p,q)}(\mathrm{diag}(1+\delta,1-\delta))$$
which is equivalent to
$$ (1+\delta)^p + (1-\delta)^q \geq 1^p + 1^q.$$
Taking $\delta$ to be small and performing a Taylor expansion, we see that this fails when $p<q$ for $\delta$ small enough.

  [1]: https://nhigham.com/2021/02/02/what-is-a-unitarily-invariant-norm/