Why can't a nonabelian group be 75% abelian? This question asks for intuition, not a proof.
An earlier question,
Measures of non-abelian-ness
was thoroughly answered by Arturo Magidin.
A paper by Gustafson1
proves that, for a nonabelian group,
the probability that two randomly selected
elements commute is at most $5/8$, a tight bound2
that also holds for a class of infinite topological groups.
One might say that a nonabelian group cannot be more than 62.5% abelian.
My question is:

Q. Is there some intuitive reason that a nonabelian
  group cannot be "nearly completely abelian" in the sense that
  the probability that two element commute approaches $1$?
  Why, intuitively, is there an upper bound less than $1$?

Gustafson's proof uses the conjugacy class equation,
and bounds on the terms of this equation.
I am seeking an underlying logic that can be conveyed without these calculations.
(Incidentally, there is no universal lowerbound on the probability, so
groups can be nearly completely nonabelian.)

1Gustafson, W. H.
"What is the probability that two group elements commute?"
American Mathematical Monthly (1973): 1031-1034.
(Jstor link.)

2"The reader may verify this bound is sharp,
by examining the nonabelian
groups of order $8$":
The dihedral group $D_8$,
and the quaternion group $Q_8$.
 A: (In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a subgroup of $G$). In other words, once the probability that $x$ commutes with elements of $G$ gets above $\frac{1}{2}$, it has to be $1$.
Also, as I mentioned, there are many papers on this topic in the literature. At the other extreme to the question, in one paper, Bob Guralnick and I proved that the probability that two elements of  a finite group $G$ commute tends to $0$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup
of $G$.
Later note: These arguments work not just with finite groups, but also whenever a group  $G$ admits a measure $\mu$  which is invariant under right (or left if preferred)  translation, making $
(G,\mu)$ a probability space. For then it is still true that if the probability that $x$ commutes with a group element gets above $\frac{1}{2}$, we must have $C_{G}(x) = G$, for otherwise $\mu(C_{G}(x)) \leq \frac{1}{2}$, as the $[G:C_{G}(x)]$ right cosets of $C_{G}(x)$ all have equal measure. Similarly, we have $[G:Z(G)] \ge 4$ if $G$ is non-Abelian, since otherwise $\mu(\langle x \rangle Z(G) ) > \frac{1}{2}$ for $x \in G \backslash Z(G)$.
