The Schur multiplier $H^2(G;{\mathbb C}^\times) \cong H^3(G;{\mathbb Z})$ of a finite group is a product of its $p$-primary parts
$$H^3(G;{\mathbb Z}) = \oplus_{ p | |G|} H^3(G;{\mathbb Z}_{(p)})$$
as is seen using the transfer. The $p$-primary part $H^3(G;{\mathbb Z}_{(p)})$ depends only of the $p$-local structure in $G$ i.e., the Sylow $p$-subgroup $S$ and information about how the subgroups of $S$ become conjugate or "fused" in $G$. (This data is also called the $p$-fusion system of $G$.)
More precisely, the Cartan-Eilenberg stable elements formula says that
$$H^3(G;{\mathbb Z}_{(p)}) = \{ x \in H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)} |res^S_V(x) \in H^3(V;{\mathbb Z}_{(p)})^{N_G(V)/C_G(V)}, V < S\}$$
One in fact only needs to check restriction to certain V above. E.g., if S is abelian the formula can be simplified to $H^3(G;{\mathbb Z}_{(p)}) = H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)}$ by an old theorem of Swan. (The superscript means taking invariants.) See e.g. section 10 of my paper linked HERE for some references.
Note that the fact that one only need primes p where G has non-cyclic Sylow $p$-subgroup follows from this formula, since $H^3(C_n;{\mathbb Z}_{(p)}) = 0$.
However, as Geoff Robinson remarks, the group $H^3(S;{\mathbb Z}_{(p)})$ can itself get fairly large as the $p$-rank of $S$ grows. However, $p$-fusion tends to save the day. The heuristics is:
Simple groups have, by virtue of simplicity, complicated $p$-fusion, which by the above formula tends to make $H^3(G;{\mathbb Z}_{(p)})$ small.
i.e., it becomes harder and harder to become invariant (or "stable") in the stable elements formula the more $p$-fusion there is. E.g., consider $M_{22} < M_{23}$ of index 23: $M_{22}$ has Schur multiplier of order 12 (one of the large ones!). However, the additional 2- and 3-fusion in $M_{23}$ makes its Schur multiplier trivial. Likewise $A_6$ has Schur multiplier of order 6, as Geoff alluded to, but the extra 3-fusion in $S_6$ cuts it down to order 2.
OK, as Geoff and others remarked, it is probably going to be hard to get sharp estimates without the classification of finite simple groups. But $p$-fusion may give an idea why its not so crazy to expect that they are "fairly small" compared to what one would expect from just looking at $|G|$...
Best Answer
Here's how I think about it: if the automorphism group of an object is abelian, this means something very strong about the object. It sort of means you can affect its structure in two different ways, in any order, and they won't affect each other. This to me hints that maybe the object consists of separate "pieces" that can only be affected independently.
The first example like this which comes to mind is a direct (commutative ring) product of finite fields of different characteristics. Each field has a cyclic automorphism group, making it very "simple", and the fields can't map into each other, meaning they can only be affected "independently."
So I'd posit that if you want to think of abelian groups as symmetry groups, you can imagine them as the symmetry groups of objects that "break apart" into "simple" pieces which can't interact with each other (intentionally vague, since I don't want to commit to a particular category).