The argument below is mainly group theoretic. I think it should be possible to give a more direct character-theoretic proof.

Let $H = O^{2}(N)$ which is cyclic of order $15$, and let $X = O^{2}(G).$ Then it is enough to prove that $X$ has no irreducible character of degree divisible by $15$, since $X \lhd G$ with $[G:X]$ a power of $2$, using Clifford's Theorem.
Note that $H \leq Z(X)$ since $H \lhd X$ and ${\rm Aut}(H)$ is a $2$-group.

Let $P$ be a Sylow $5$-subgroup of $X$. Then $P$ is Abelian and $P \cap X^{\prime} \cap Z(X) \leq P^{\prime} = 1$ by elementary transfer. A similar argument works for the prime $3$. Hence $H \cap X^{\prime} =1$.

Now $XN/N \lhd G/N$ so either $X \leq N$ or $XN = G$. But if $X \leq N$ then $G$ is solvable, which is not the case. Hence $XN = G$ and $[G:X] = 2$.

Let $M$ be the terminal member of the derived series for $G$. Then $M \leq X$, so $M \leq X^{\prime}$. Hence $M \cap H = 1$ and $|M| \leq 60.$ Thus $M \cong A_{5}$ and $X \cong A_{5} \times H$, so all irreducible characters of $X$ have degree $1,3,4$ or $5$.