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.

1more comment