On the integrals $\int_0^\infty [1-x^n \operatorname{arccot}^n(x)] \, \mathrm{d} x$

Cool problem. Both proposed asymptotic forms are correct. I'm only going to do a first order calc for both, and for $J_n$ I'll say where the proof needs some work. For $I_n$ use the expression given by Arathoon's first note. Observe that $t\cot(t)$ is unimodal decreasing and at $t=0$ has the value 1. Thus by raising it to a high power we expect it to become more sharply peaked. In fact it is gaussian because $$ t\cot(t) = 1-t^2/3-t^4/45... \sim \exp(-t^2/3)(1-7/90t^4 + ...) $$ Therefore $$I_n = \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-(t\cot(t))^n \big) \sim \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-\exp{(-n\,t^2/3)} \big).$$ Because $n \to \infty$ and because of the $\csc^2t,$ it is seen that the most important region is near $t=0.$ The trick now is to recognize that $\tan^2t = t^2 + O(t^4).$ Replace $t^2$ in the exponential with $\tan^2t$ because it just so happens (Mathematica knows it too) that $$ \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-\exp{(-a\,\tan^2t)} \big)=\sqrt{a \pi}.$$ With $a=n/3,$ we indeed have $I_n \sim \sqrt{n \pi /3}.$

For $J_n$ I used a technique call $\textit{depoissonization}.$ Make an exponential power series, $$\sum_{n=0}^\infty \frac{y^n}{n!}J_n = \sum_{n=0}^\infty\frac{y^n}{n!} \int_0^{\infty}(1-t\cot^{-1}(t))^n\, dt = e^y\int_0^{\infty}\exp{(-y\,t\,\cot^{-1}{t} )} \, dt. $$ Since $t\cot^{-1}t$ is monotonically increasing, all we need is the behavior near $0$ to asymptotically estimate the integral. In fact, $t\cot^{-1}t = \pi\,t/2 +O(t^2).$ Using the first term we therefore find $$J(y) := e^{-y}\sum_{n=0}^\infty \frac{y^n}{n!}J_n \sim \int_0^{\infty} \exp{(-\pi\,y\,t/2)} \, dt = \frac{2}{\pi\,y}.$$ By depoissonization, $J_n \sim J(n) =2/(\pi n)$ as long as the next term in the sequence is smaller than this first term. I don't intend to show that. One thing that makes this problem interesting is that in fact it can be shown that $I_n = \sum_{k=1}^n(-1)^{k+1}\binom{n}{k}J_k$ so that $I_n$ and $J_n$ are binomial transforms. Putting in the first order asymptotic for for either $I_n$ or $J_n$ will $\textit{not}$ give you the asymptotics of its transform.