The sum (with multiplicity) of the cubes of irreducible character degrees of a finite group Throughout $G$ is a finite, non-abelian group.
$\DeclareMathOperator\Irr{Irr}\DeclareMathOperator\AD{AD}\DeclareMathOperator\cp{cp}\newcommand\card[1]{\lvert#1\rvert}$
Let $\Irr(G)$ be the set of irreducible characters (working over the complex ground field). Let
$$ \AD(G) = \card G^{-1} \sum_{\chi\in\Irr(G)} (d_\chi)^3 \;. $$
(The invariant $\AD(G)$ arose under a different name in work of Johnson on amenablity constants of Fourier algebras, and is a special case of something more general that can be defined for countable virtually abelian groups in terms of Plancherel measure for such groups. But this question is only concerned with character theory of finite groups.)
Question. If $\AD(G)\leq 2$, does this imply that $d_\chi\leq 2$ for all $\chi\in\Irr(G)$?
Here is my naive reasoning to support my guess that the answer is positive. Let $\Irr_n(G)=\{\chi\in\Irr(G) \mid d_\chi =n\}$. Then
$$
2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) =
\sum_{\chi\in\Irr_1(G)} 3 + \sum_{\chi\in\Irr_2(G)} 12 + \sum_{n\geq 3} \sum_{\chi\in\Irr_n(G)} (d_\chi)^3 $$
and therefore, since $\card G=\sum_{n\geq 1} \sum_{\chi\in \Irr_n(G)} (d_\chi)^2$,
$$
2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) \geq 3\card G.
$$
If we impose the condition that $\AD(G)\leq 2$, this yields
$$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} \geq \card G $$
which seems difficult to achieve if there are irreps of degree $\geq 3$.
EDIT: the original version of this post asked if there existed any $G$ with irreps of degree $\geq 3$ that satisfied this inequality, with the naïve hope that no such $G$ existed. Victor Ostrik has pointed out in a comment that an extraspecial $2$-group of order $32$ has $16$ linear characters and a single nonlinear character which has degree $4$, so that the inequality above is indeed satisfied. On the other hand the AD of this group is $(16+64) / 32 = 5/2$.

Some other background facts which may or may not be useful.

*

*If $H\leq G$ then $\AD(H)\leq\AD(G)$. I am not sure if this has a simple proof just using the definition above, but it follows from the alternative description in tems of "amenability constants of Fourier algebras".


*There is some relationship between $\AD(G)$ and the commuting probability
$$ \cp(G) = \frac{\# \hbox{conjugacy classes of $G$}}{\card G} = \frac{\card{\Irr(G)}}{\card G} $$
which shows that for the former to be “small” the latter should be “large”. To be precise, note that Hölder with conjugate exponents 3 and 3/2 gives
$$
\sum_{\chi\in\Irr(G)} (d_\chi)^2 \leq \left( \sum_{\chi\in\Irr(G)} 1^3 \right)^{1/3}  \left( \sum_{\chi\in\Irr(G)} (d_\chi)^3\right)^{2/3}
$$
so that
$$ 1 \leq \cp(G) \AD(G)^2. $$
Moreover, equality is strict since we assume $G$ is non-abelian. In particular, if $\AD(G)\leq 2$ this implies $\cp(G) > 1/4$. Can this be leveraged, perhaps by using the good behaviour of $\cp$ with respect to quotients? I had a quick look in the 2006 paper On the commutating probability in finite groups of Guralnick–Robinson and I suspect that results there could be useful, but as I am not a specialist in finite group theory I couldn't see how to make progress.
 A: Here are a few remarks which may be helpful. I may be able to say more later.  The condition that $\operatorname{cp}(G) > \frac{1}{4}$ already severely restricts the possibilities for $G$. If $G$ is a finite group which does not have a normal Sylow $p$-subgroup for a prime $p$, then it is proved in (GR 2006) mentioned in the question, that we have $\operatorname{cp}(G) \leq \frac{1}{p}$. Hence $G$ has a normal Sylow $p$-subgroup for each prime $p >3$. Also, it follows from the results of that paper that $G$ has an Abelian Sylow $p$-subgroup for each prime $p > 3$. It follows that $G$ has an Abelian normal subgroup $A$ with $[G:A]=2^{a}3^{b}$ for non-negative integers $a$, $b$. Notice that such a group $G$ is solvable.
However, if $G$ is the direct product of an extra-special group of order $8$ and an extra-special group of order $27$, then we have $\operatorname{cp}(G) = \frac{55}{216} > \frac{1}{4}$.
Nevertheless, it is true that $\operatorname{AD}(X \times Y) = \operatorname{AD}(X)\operatorname{AD}(Y)$, and for $G$ above we obtain $\operatorname{AD}(G) = \frac{3}{2} \times \frac{7}{3} = \frac{7}{2} > 2$.
Later edit: It is also proved in (GR 2006) that when $G$ is solvable  (which is the case now) we have $\operatorname{cp}(G)^{2} \leq \frac{\operatorname{cp}(F)}{[G:F]}$, where $F = F(G)$, the (unique) largest nilpotent normal subgroup of $G$.
Since $\operatorname{cp}(G) > \frac{1}{4},$ this forces $[G:F] \leq 12$ if $F$ is Abelian and $[G:F] \leq 9$ if $F$ is non-Abelian. This is because $G/F$ is a $\{2,3\}$-group , while we have $[G:F] < 16$ if $F$ is Abelian, and $[G:F] < 10$ if $F$ is non-Abelian ($\operatorname{cp}(F) \leq \frac{5}{8}$ in the latter case, by a Theorem of W.Gustafson).
So far, this does not seem to resolve your problem, but it does at least suggest a small bound on the largest irreducible character degree if
$\operatorname{AD}(G) \leq 2$. For it is a theorem of N. Ito that the degree of an irreducible character of a finite group $G$ divides the index of any Abelian normal subgroup of $G$ (in fact, it is mentioned in Isaacs' book on Character Theory that each irreducible character of $G$ has degree at most $n$ if $G$ has an Abelian subgroup of index $n$, even if the Abelian subgroup is not normal).
If $F$ is Abelian, we may conclude that every irreducible character of $G$ has degree at most $12$.
In fact, if $O_{2}(G) = O_{2}(F)$ is Abelian, but $F$ is non-Abelian, we obtain $[G:F] \leq 6$ because $\operatorname{cp}(F) \leq \frac{11}{27}.$ In that case,
we must have $G/F$ Abelian, for otherwise $G/F \cong S_{3}$ and
$\operatorname{cp}(G) \leq \operatorname{cp}(G/F)\operatorname{cp}(F) \leq \frac{11}{54} < \frac{1}{4},$ a contradiction.
By the way, detailed analysis may seem messy, but Abelian normal subgroups can't be ignored, because we always have $\operatorname{AD}(G \times H) = \operatorname{AD}(G)$ whenever $G$ is a finite group and $H$ is a finite Abelian group.
Later edit: I have been looking up some results in the other direction, where a conjecture of D. Gluck seems very relevant. For example, Alex Moreto and Tom Wolf have proved that if $G$ is a finite solvable group, then $G$ has an irreducible character whose degree is at least $[G:F(G)]^{\frac{1}{3}}$.
Hence the only finite solvable groups $G$ which have no irreducible character of degree greater than $2$ have $[G:F(G)] \leq 8.$
Also, Cossey, Halasi, Maroti and Nguyen have proved that if $G$ is a finite solvable group with $|F(G)|$ of order coprime to $6$, then $G$ has an irreducible character whose degree is at least $[G:F(G)]^{\frac{1}{2}}$. If the conjecture of Gluck is true, then the restriction that $|F(G)|$ has order coprime to $6$ could be removed.
In any case, at present it is known that the only finite solvable groups $G$ which have $|F(G)|$ of order coprime to $6$ and no irreducible character of degree greater than $2$ have $[G:F(G)] \leq 4.$
To conclude: if your question has a positive answer, then it must be the case that any finite group $G$ with $\operatorname{AD}(G) \leq 2$ has $[G:F(G)] \leq 8$ (recall that such a group $G$ has been shown to be solvable). Note that this is a little stronger than I managed to prove above.
