Original problem: suppose a group $G$ has irreducible complex representations of dimensions $1,1,2,3$, and $d$. Find $d$.
Using some basic dimension counting, we immediately get $d=3$ or $d=15$. The goal is to eliminate $d=15$.
If $d=15$, then $|G|=1^2+1^2+2^2+3^2+15^2=240$. Since $G$ has two $1$-dimensional irreducible representations, $[G:G']=2$, and $G'$ is normal. Thus $G'$ is a union of conjugacy classes. But $G'$ has elements of order $1$, $2$, $3$, and $5$, so it contains at least $4$ conjugacy classes. The only possibility is that $G\setminus G'$ is a single conjugacy class of order $120$.
At this point, it is driving me crazy that I can't find an easier contradiction. It seems absurd that $G$ can have $240$ elements with half of them in a single conjugacy class. I've included a solution I found which I am comfortable with, but which seems a bit convoluted to me.
Solution: For any $x\in G\setminus G'$, the centralizer $C_G(x)$ has order $2$, so $x^2=1$. Then $C_G(x)$ extends to a Sylow $2$ subgroup $P$. The center of $P$, $Z(P)$, is nontrivial, and $Z(P)\subseteq C_G(x)$, so $C_G(x)=Z(P)$. But if $x$ is central in $P$, then $|C_G(x)|\ge |P|=16$.
A different approach is to use a list of all $8$ groups with $5$ conjugacy classes. According to GAP the orders of these groups are $\ 5,8,14,20,21,24,60. \ $ Because the order of the group must be $\ n := 1^2+1^2+2^2+3^2+d^2 = 15+d^2 \ $ and we can eliminate groups with prime order, that leaves $\ 16,24,40,51,64,\dots \ $ and so that implies $\ n = 24 \ $ and GAP has only $\ S_4 \ $ in the first list.