EDITED IN RESPONSE TO COMMENTS BY DAVID SPEYER AND F. LADISCH: An example, which is effectively definitive, is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity elements acts without the eigenvalue $1$. Such groups have cyclic Sylow $p$-subgroups for all odd primes $p,$ and cyclic or generalized quaternion Sylow $2$-subgroups, properties which also occur in Will Sawin's answer . This is a very restricted class of groups. For example, the only perfect Frobenius complement is ${\rm SL}(2,5).$ In any case, if $G$ is a Frobenius complement, and $\chi$ is a faithful complex irreducible character such that $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle =0$ for each non-identity cyclic subgroup $H$ of $G$ (and such a $\chi$ must exist), then $\chi$ is not a constituent of any permutation character induced from the trivial character of a non-identity subgroup of $G.$

Let me justify that Frobenius complements are just those groups with an irreducible complex representation where every non-identity element acts without eigenvalue $1$ . Recall that a Frobenius group $G$ has the form $G = KH,$ where $K \lhd G$ and $H \cap K = 1,$ and, furthermore, $H \cap H^{g} = 1$ for all $g \in G \backslash H.$

Notice then that $|K| \equiv 1$ (mod $|H|$), so that ${\rm gcd}(|K|,|H|) = 1.$ Also, we certainly have $C_{G}(h) \leq H$ whenever $h$ is a non-identity element of $H$ since $h \in H \cap H^{c}$ for all $c \in C_{G}(h).$

Let $V$ be a minimal non-identity $H$-invariant subgroup of $K.$ Then using Thompson's theorem that a Frobenius kernel is nilpotent, (which is actually overkill here, since by general properties of coprime automorphism groups to be found in Gorenstein's book "Finite Groups", for example, $H$ normalizes a Sylow $q$-subgroup of $V$ for each prime divisor $q$ of $|V|$), we see that $V$ is an elementary Abelian $p$-group for some prime $p$. Then $V$ is a faithful $FH$-module, where $F = {\rm GF}(p).$ Sin $p$ does not divide $|H|$, $V$ "lifts" to a complex representation, and by general (indeed the defining) properties of Brauer characters, it is still the case that each non-identity element of $H$ acts without eigenvalue $1.$ The "lifted" repesentation need not be irreducible as a complex representation, but its irreducible components all have the property that each on-identity element of $H$ acts without eigenvalue $1$ on them (and each is faithful.

Conversely, if $H$ is a finite group which has a complex irreducible character $\chi$ which does not contain the trivial character on restriction to any non-identity cyclic subgroup of $H,$ then a complex representation of $\chi$ may be reduced $mod $p$)$ for any prime $p$ which does not divide $|H|$ to afford a $kH$-module $W$ on which each non-identity element of $H$ acts without non-zero fixed points, where $k$ is algebraically closed of characteristic $p.$ Then $W$ may be realised over a finite field, and the sum of its distinct Galois conjugates may be realised over ${\rm GF}(p),$ say by module $V$ over ${\rm GF}(p).$ It is still te case that each non-identity element of $H$acts without non-trivial fixed points on $V,$ so the semidirect product $VH$ is a Frobenius group with kernel $V$ and complement $H.$
Frobenius complements are precisely the groups which have an irreducible character $\chi$ which does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$ for any non-trivial subgroup $H$ of $G.$ For if $\chi$ is such a character, then ${\rm Res}^{G}_{H}(\chi)$ has no trivial constituent for each non-trivial subgroup $H$ of $G$, in particular for each non-trivial cyclic subgroup of $G.$ Hence each non-identity element of $G$ acts without eigenvalue $1$ in any complex representation affording $\chi.$ Conversely, if each non-identity element of $G$ acts without eigenvalue $1$ in a representation of $G$, then there is an irreducible representation $\sigma$ with that property, and if $\sigma$ affords character $\chi,$ then $\chi$ does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$for any non-trivial subgroup $H$ of $G.$
There are examples of non-Abelian Frobenius complements of odd order: for example, let $G = \langle x,y : x^{9} = y^{7} = 1, x^{-1}yx = y^{2} \rangle.$ Note that $G$ has an irreducible character $\chi$ of degree $3$ such that $x^{3}$ acts, as a non-identity scalar matrix, so that no non-identity $3$-element of $G$ has eigenvalue $1$ in the associated representation, while also each non-identity power of $y$ has three different primitive $7$-th roots of unity as its eigenvalues in the associated representation.
However, it is not true that if a finite group of odd order has all its Sylow subgroups Abelian, then it is a Frobenius complement: for example, anon-Abelian group of order $21$ is not a Frobenius complement (though it is a Frobenius GROUP!)