This isn't really an answer as much as it is an "expanded" comment.
Consider, for integer $a$, $$P_n(x) = \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n = \sum_{m=0}^n \binom{n}{m} \binom{n+m}{m} (-ax)^m$$
Given $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) (x^x(1-x)^{1-x})^q - 1\right) \ dx$$
We have
$$I_n \leq \int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx$$ where $a_q = \max_{x\in (0,1)}\{(x^x(1-x)^{1-x})^q\}$. Furthermore, we have
$$\left|\int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx \right|= \left|\int_0^1 \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n\left( \sin(\pi q x) a_q - 1\right) \ dx\right|$$
$$= \left|\frac{1}{a^{n+1}n!} \int_{(0,1)\cup(1,a)} \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$
$$\leq \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{1}{n!a^{n+1}}\int_1^a \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$
Let $$S_n = \int_1^a \left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right)\frac{d^n}{d x^n} x^n (1-x)^n \ dx$$
So that we have
$$ = \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right|$$
Now, observing the bound in question, applying Theorem 4, and letting $A = C - \gamma + o(1)$, we have
$$\left| (n+q+2)!_\mathbb{P} (n+1) I_n \right|<\left|e^{An} e^{q+1} n^n \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}n^{q+1}(n+1)\left(\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right) \right| $$
If we ignore the $S_n$ term, we have that
$$\left| e^{q+1} \frac{n^n}{e^n n!} \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}\frac{n^{q+1}(n+1)}{b^n}\left(\frac{e^{A+1}\pi bq}{4a^2}\right)^n \frac{a_q}{a} \right|$$
where $b > 1 $. If we consider $a$ such that $ a^2 > \frac{e^{A+1}\pi bq}{4}$, and applying Stirling's approximation to the left-most term (-ish), for large $n$, then the whole expression above tends to $0$. Now, it is left to consider the $S_n$ term, though I have a bad feeling about it. :/
Best Answer
The expected value is $2$, simply because almost all real numbers have irrationality measure $2$. Note that this is "almost all" in the sense of measure theory. In the sense of topological category, it's the other way around: the set of reals of measure $\infty$ is residual. As I pointed out recently, it's a bit mysterious why measure theory makes better predictions than topological category for "naturally defined" numbers like $e$ or $\pi$, but experimentally, this appears to be the case, as the upper bound on the irrationality measure of $\pi$ is finite.