Finite groups with a character having very few nonzero values?
For n=2, your question is addressed in:
Gagola, Stephen M., Jr. "Characters vanishing on all but two conjugacy classes." Pacific J. Math. 109 (1983), no. 2, 363–385. MR721927 euclid.pjm/1102720107
In it, he shows that if such a nearly-zero character exists, it is unique, and the unique faithful irreducible character of G (similar to the extra-special 2 and 3 groups mentioned by Kevin Buzzard). A doubly transitive Frobenius group has such a character, but the non-solvable doubly transitive Frobenius groups are in short supply (five or so). The possible forms of more general non-solvable examples are restricted in Theorem 5.6 (basically SL2).
Berkovich and Zhmud have results for more general cases. See chapter 27 of their book on character theory (volume 2) which answers a broader question for n=2 and n=3:
Which groups have all irreducible non-linear characters taking on only n distinct values?
The list of groups is fairly short, but I haven't had time to verify it; they are all far from simple, usually with a normal Sylow and nilpotent quotient. They have a result for n=4 too, on p. 239, but it is much less complete.
Since your character is nonzero except on n classes, it can take on at most n nonzero values. It is fairly likely to be faithful, since every class contained in the kernel is a wasted class that cannot be zero. This allows some of the lemmas to be used. If Gagola's case is indicative, then the "all" versus "one" will not actually be a huge difference, at least mod the kernel of your single character.
Here are the paper references:
Berkovich, Yakov; Chillag, David; Zhmud, Emmanuel. "Finite groups in which all nonlinear irreducible characters have three distinct values." Houston J. Math. 21 (1995), no. 1, 17–28. MR1331241
Zhmudʹ, È. M. "On finite groups, all of whose irreducible characters take at most two nonzero values." Ukraïn. Mat. Zh. 47 (1995), no. 8, 1144–1148;translation in Ukrainian Math. J. 47 (1995), no. 8, 1308–1313 (1996) MR1367729 DOI:10.1007/BF01057720
Berkovich and Zhmud also pose an exercise that if all the irreducible characters of degree coprime to p take on only 3 values, then the group is p-solvable. The same is true if "coprime to" is replaced by "divisible by". This is page 238.
For the case of n=2 see Gagola's paper "Characters vanishing on all but two conjugacy classes" MR0721927. Some improvements on Gagola's results can be found in Kuisch and van der Waall's papers "Homogeneous character induction [I and II]" MR1172440/MR1302857 and in my paper "Groups with a Character of Large Degree".
See Theorem 4.1 in my paper for what I think is the strongest result in this direction (as far as I know):
Gagola showed that the character of $V$ vanishes on all but one nontrivial conjugacy class if and only if there exists a subgroup $N$ of $G$ such that $N$ acts trivially on each simple $\mathbb{C}[G]$-module except $V$.
Let $G$ be a group of order $n$ and $V$ a simple $\mathbb{C}[G]$-module of dimension $d$. Define $e$ such that $n = d(d+e)$. Assume that there exists a normal subgroup $N \neq > \{1\}$ such that $N$ acts trivially on each simple $\mathbb{C}[G]$-module other than $V$. Let $x$ be a nontrivial element of $N$ and $C$ be the centralizer of $x$ in $G$. Then there exist a prime number $p$, a positive integer $k$ and a non-negative integer $m$ such that:
- $N$ is elementary abelian of order $p^k$,
- $C$ has order $p^k e^2$, and $d = e(p^k-1)$, and $n = e^2 p^k (p^k-1)$,
- $C$ is a Sylow $p$-subgroup of $G$ and $e = p^m$,
- if $H$ is any group such that $N \subseteq H \subseteq C$ and $|H| > > p^{k+m}$, then $N \subseteq [H,H]$.
The above conditions are also sufficient, see Thm 4.3
For n bigger than 2 there may be some results in MR1487363 Chillag "On zeroes of characters of finite groups."