By the standard classification of unitarily invariant norms (see e.g., this blog post, 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 $y^p+|x|^q$ 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 $p \geq q$. As $S$ is symmetric across the diagonals $x=y$ and $x=-y$, we conclude that 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 sees that if one wishes $N_{(p,q)}$ to be a norm, 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.