Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$. 
Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.

I found the result using a trick (using $e^x = \tan(\theta/2)$ as mentioned here) but I wanted to know if there isn't another less astute approach.

There were a variety of approaches presented on the page of the embedded link in the OP.
Here is yet another way forward.

Enforce the substitution $e^x\mapsto x$.  Then, we see that
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx&=2^n\int_0^\infty \frac{1}{(x+x^{-1})^n}\frac1x \,dx\\\\
&=2^n\int_0^\infty \frac{x^{n-1}}{(x^2+1)^n}\,dx\tag1
\end{align}$$

Next, enforcing the substitution $x\mapsto x^{1/2}$ in $(1)$ reveals
$$\begin{align}
2^n\int_0^\infty \frac{x^{n-1}}{(x^2+1)^n}\,dx&=2^{n-1}\int_0^\infty \frac{x^{n/2-1}}{(x+1)^n}\,dx\\\\
&=2^{n-1}B\left(\frac{n}{2},\frac{n}{2}\right)\\\\
&=B\left(\frac{n}{2},\frac{1}{2}\right)\tag2
\end{align}$$
where $B(x,y)$ is the Beta Function.  Note that the result in $(2)$ agrees with the result reported on the referenced page.


The equivalence of $2^{n-1}B\left(\frac n2,\frac n2\right)=B\left(\frac n2,\frac12\right)$ can be established using the relationship between the Beta Function and the Gamma Function, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ along with Legendre's Duplication Formula of the Gamma Function, $\Gamma (z)\Gamma (z+1/2)=2^{1-2z}\Gamma(1/2)\Gamma(2z)$ with $z=n/2$.


For large $n$, we can use the relationship between the Beta Function and the Gamma Function, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, along with Stirling's Approximation to arrive at the asymptotic approximation
$$B\left(\frac{n}{2},\frac{1}{2}\right)\sim\sqrt{\frac{2\pi}{n}}$$
Therefore, we see that for large $n$ the integral of interest has the asymptotic approximation
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\sim \sqrt{\frac{2\pi}{n}}}$$


APPENDIX:  SIMPLE ESTIMATES TO OBTAIN THE ASYMPTOTIC EXPANSION
It is straightforward to show that the hyperbolic cosine satisfies the inequalities
$$1+\frac12x^2\le \cosh(x)\le e^{x^2/2}\tag{A1}$$
Hence, we have from $(A1)$
$$\int_{-\infty}^\infty e^{-nx^2/2}\,dx \le\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\le \int_{-\infty}^\infty \frac{1}{\left(1+\frac12 x^2\right)^n}\,dx \tag{A2}$$
The value of the integral on the left-hand side of $(A2)$ is $\sqrt{\frac{2\pi}n}$ while the value of the integral on the right-hand side of $(A2)$ can be found by making the substitution $x\mapsto \sqrt 2\tan(x)$.  To wit, we see that
$$\int_{-\infty}^\infty \frac1{\left(1+\frac12 x^2\right)^n}\,dx=2\sqrt{2}\int_0^{\pi/2}\cos^{2n-2}(x)\,dx\tag{A3}$$
The integral on the right-hand side of $(A3)$ is Wallis' Integral with exponent $2n-2$.
It is straightforward to show (See Here) that $\lim_{n\to \infty}\sqrt{n}\int_0^{\pi/2}\cos^{2n-2}(x)\,dx=\frac{\sqrt{\pi}}{2}$ whence application of the squeeze theorem reveals
$$\lim_{n\to\infty}\sqrt{n}\int_{-\infty}^\infty \frac1{\cosh^n(x)}\,dx=\sqrt{2\pi}$$
from which we find the asymptotic approximation
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\sim \sqrt{\frac{2\pi}{n}}}$$
as expected!
