[This answer contradicts Jim Humphreys's, and I'm guessing his is right and mine is wrong, but so far I can't see why.]
By Frobenius reciprocity, a representation is induced from the trivial representation of the subgroup $H$ if and only its restriction to $H$ includes the trivial representation. So every non-cyclic abelian group has this property, because an irreducible representation is one-dimensional, so factors through a map to a cyclic group, so has a kernel.
Moreover, this implies that if $H\subset G$ and $H$ has this property, then $G$ does as well, so a sufficient condition is that a group have an abelian subgroup that's not cyclic.
This resolves the case for $GL_n(\mathbb F_p)$, $n>1$ and $p>2$ (the diagonal subgroup), and $n>2$ and $p>3$ (an abelian subgroup of the $2$-Sylow subgroup). n=2, p=2 is just $S_3$, which manifestly has this property. (Showing that this sufficient condition is not necessary) So $GL_n$ for $n>1$ has this property. $GL_1$ does not, since it is always cyclic. (For cyclic groups, every induced representation fails to be faithful, so take a faithful 1-dimensional irrep.)
$A_n$ for $n\geq 4$ contains the Klein four subgroup of $A_4$ and so has this property. $A_3$ is cyclic and so does not.
I do not know an easy-to-check necessary condition.