Here's half a proof:

The answer should be e=3. You've showed that's always realizable, so all I need to do is exhibit infinitely many n so that any group with an n-dim irrep has dimension at least c n^3, for some fixed constant c < 1/2. All the n's I exhibit will be prime, so let n=p. I also want that (p-1)/2 is prime.

Assume for contradiction a group G with #G smaller than p^3 and a p-dimensional irrep. Consider a p-Sylow subgroup of G. Clearly it has size p or p^2. In particular there are only three options: Z/p, Z/p^2, and Z/p x Z/p.

If #G_p is p^2, then the number of Sylow p subgroups is congruent to 1 and divides G/p^2 < p. Hence G_p is a normal subgroup. The dimension of any irrep of a group always divides the index of any normal abelian subgroup. Hence G has size at least p^3.

So G_p is Z/p. If G_p were normal, then by the above fact p would divide its index in G, which is nonsense. So G_p has kp+1 conjugates. So the normalizer of G_p has index kp+1. Now we use that (p-1)/2 is prime to conclude that either G_p is central in its normalizer, or else the normalizer has size at least p(p-1)/2. In the latter case the whole group has size at least p(p-1)(p+1)/2.

Therefore G_p=Z/p is central in its normalizer so by the Burnside normal complement theorem it has a normal complement and G = N \semidirect Z/p for some normal subgroup N. By the "transport de structure" theorem from Serre, any irrep of G is either induced from a representation of a proper subgroup H containing N, or else its restriction to N is isotypic. Since the index of N is p which is the dimension of the irrep, in the first case the rep is induced from a 1-dim irrep of N. In the latter case, since N has no p-dimensional irreps, the restriction of to N is p copies of a 1-dim irrep. But by Frobenius reciprocity the induction of this 1-dim rep would contain p copies of our irrep, which is ridiculous. So our irrep is induced from a 1-dim rep of N.

Now we can replace G with G/[N,N] (which makes sense because a characteristic subgroup of a normal subgroup is normal), and we still have a p-dim rep of G (given by the same induction), and we've shrunk the size. So we can assume that N is abelian. To finish the proof we need to use the right version of the number theory argument David suggested.

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