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 $$ S := \{ (x,y): \max(|x|,|y|)^p + \max(|x|,|y|)^q \leq 1 \}$$ is convex (certainly it is symmetric and has non-empty interior). 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 $(a,a)$ of the set on the upper right edge of the sector, where $a>0$ is the unique positive solution to the equation $a^p+a^q=1$, implicit differentiation of the condition $y^p + |x|^q = 1$ reveals that the boundary of this set has a slope of $-\frac{q a^{q-1}}{p a^{p-1}}$ at this point, which is less than or equal to $-1$ iff we have the condition $$ p a^p \geq q a^q. \quad (1)$$ By symmetry along the y-axis, at the opposite endpoint $(-a,a)$, the slope is at most $+1$ iff the same condition (1) holds. As $S$ is symmetric across the diagonals $x=y$ and $x=-y$, we conclude that the entire set $S$ is convex iff (1) holds, and so $N_{(p,q)}$ is a matrix norm iff (1) holds also.
To obtain an explicit counterexample to the triangle inequality when (1) fails: observe that $\mathrm{diag}(a,a)$ is in the convex hull of the matrices $\mathrm{diag}(a+\delta,a-\delta)$, $\mathrm{diag}(a-\delta,a+\delta)$ for any $0 < \delta < a$, so by symmetry and the triangle inequality one sees that if one wishes $N_{(p,q)}$ to be a norm, one needs $$ N_{(p,q)}(\mathrm{diag}(a,a)) \leq N_{(p,q)}(\mathrm{diag}(a+\delta,a-\delta))$$ which is equivalent to $$ (a+\delta)^p + (a-\delta)^q \geq a^p + a^q = 1.$$ Taking $\delta$ to be small and performing a Taylor expansion, we see that this fails when (1) fails for $\delta$ small enough.
To see when (1) holds, we make the change of variables $a^p = \theta$, $a^q = 1-\theta$ for some $0 < \theta < 1$, then $$ \frac{p}{q} = \frac{\log \theta}{\log(1-\theta)}$$ and the condition (1) rearranges to $$ \theta \log \theta \leq (1-\theta) \log (1-\theta).$$ This turns out to hold precisely when $\theta \geq 1/2$, or equivalently $p \geq q$ (the difference $\theta \log \theta - (1-\theta) \log (1-\theta)$ is odd, vanishes at $\theta=1/2,1$, and concave for $1/2 \leq \theta \leq 1$).