Interesting question. We may start from the definition of the Beta function:
$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$
and rewrite it when $m\to 2m$:
$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$
$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$
Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series
$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$
Hence
$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$
And easily write:
$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$
Calling now
$$2m-1+k = a ~~~~~~~~~~~ n = b$$
We notice that the integral is well known:
$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$
Then we end up with the partial result (re-expanding $a$ and $b$):
$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$
That series does exist and it does converge to a known result:
$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
What you end up with is a sort of recursive relation for the Beta function:
$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
BUT
The above expression can be simplified!
$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.
$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.
Seems like that this is the only "duplication formula" for the beta function.
(Also, I found nothing on reviews or literature).
I disagree with correct solution in the OP and will present my own analysis for your scrutiny. First, let's reduce the clutter and set $a=E_n$. We seek to find $a$ such that
$$2\int_0^{a^{1/4}}\sqrt{a-x^4}dx=\left(n+\frac{1}{2}\right)$$
Now let $x^4=at$ or $x=(at)^{1/4}$, then
$$
dx=\frac{a^{1/4}}{4}t^{-3/4}dt\\
\sqrt{a-x^4}=\sqrt{a}\sqrt{1-t}\\
x=a^{1/4}\to t=1
$$
Substituting and rearranging we get
$$\frac{a^{3/4}}{2}\int_0^1 t^{-3/4}(1-t)^{1/2}dt=\left(n+\frac{1}{2}\right)$$
Introducing the complete beta function,
$$B(\nu,\mu)=\int_0^1 t^{\nu-1}(1-t)^{\mu-1}dt=\frac{\Gamma(\nu)\Gamma(\mu)}{\Gamma(\nu+\mu)}$$
Clearly, $\nu=1/4$ and $\mu=3/2$ and we can then show that
$$a=\left[ \frac{2\left(n+\frac{1}{2}\right)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$$
At the OP's suggestion, we can substitute
$$
\Gamma(7/4)=(3/4)\Gamma(3/4)\\
\Gamma(3/2)=\sqrt{\pi}/2\\
\Gamma(1/4)\Gamma(3/4)=\pi\sqrt{2}
$$
and demonstrate that
$$E_n=\left[ \frac{2\Gamma(3/4)^2\left(n+\frac{1}{2}\right)}{\pi\sqrt{2\pi}}\right]^{4/3}$$
So, it would appear that the original correct solution was missing a factor of $\pi$ in the denominator.
The solution I present here has been validated numerically for $n\in\mathbb{R^+}$, i.e., not just $n\in\mathbb{Z}$.
Best Answer
For $a$ or $b$ going to infinity, take the limit inside the integral and see that the integrand vanishes, so the integral term goes to zero. Same with when both go to infinity.
In more details:
Consider $$ B(a,b) = \int_0^1 x^{a -1}(1-x)^{b-1}\,d x~ \tag 1. $$
What happens when $a$ goes to infinity? This is equivalent to asking what happens in $\lim_{n\to\infty} \int_0^1 f_n(x)d x $ where, $f_n(x)=x^{a_n -1}(1-x)^{b-1}\,$ and $a_n\to\infty$.
Now, it would be really convenient if we could take the limit inside the integral, and it so happens that we can totally do that. We need to use a version of what is called the Monotone Convergence Theorem.
Once we take the limit inside, for any $x\in (0,1)$, for any $b$, $f_n(x)$ decreases monotonically to zero as $n\to\infty$. So, basically $f_n$ converges pointwise to the zero function. Now integration of the zero function of the interval is definitely zero.
Check this out if you want to know the technical details of the theorem we used to interchange limit and integration.
Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions
Same happens for the other limits you want to know.