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
We introduce the analytical continuation of Beta-function $\Bigl (B(\gamma,\alpha)=\int_0^1s^{\gamma-1}(1-s)^{\alpha-1}ds$, if $\gamma, \alpha >0\,\Bigr)$ for $\boldsymbol{negative}$ $\gamma\in(-1,0)$: $$B(\gamma,\alpha)=-\frac{1}{(\exp(2\pi{i}\alpha)-1)(\exp(2\pi{i}\gamma)-1)}\oint_Ps^{\gamma-1}(1-s)^{\alpha-1}ds$$
where $P$ is Pochhammer contour in the complex plane. It can be proved that analytically continued Beta-function is expressed in the usual way in terms of Gamma-function: $B(\gamma,\alpha)=\frac{\Gamma(\gamma)\Gamma(\alpha)}{\Gamma(\gamma+\alpha)}$. This expression is valid for all complex $\alpha, \gamma$.
It can also be shown that for $\gamma\in(-1;0)$ and $\alpha>0$ $$B(\gamma,\alpha)=\lim_{r\to0}(\int_r^1s^{\gamma-1}(1-s)^{\alpha-1}ds+\frac{r^\gamma}{\gamma})$$ (You can have a look, for example, here - I apologise for a clumsy explanation, but I did not manage to find this theme in an appropriate form in open sources).
The rest is straightforward: $$J(\gamma,\alpha,\alpha')=\int_0^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds$$$$=\lim_{r\to0}\int_r^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds=\lim_{r\to0}\left(B(\gamma,\alpha)-\frac{r^\gamma}{\gamma}-B(\gamma,\alpha')+\frac{r^\gamma}{\gamma}\right)$$ $$J(\gamma,\alpha,\alpha')=B(\gamma,\alpha)-B(\gamma,\alpha')$$ In your case $$I=\frac12\int_0^1\left(-1+(1-s)^{-\frac12}\right)\cdot s^{-\frac54}ds=\frac{1}{2}\Big(B\big(-\frac{1}{4};\frac{1}{2}\big)-B\big(-\frac{1}{4};1\big)\Big)$$ $$=\frac{1}{2}\bigg(\frac{\Gamma\big(-\frac{1}{4}\big)\Gamma\big(\frac{1}{2}\big)}{\Gamma\big(\frac{1}{4}\big)}-\frac{\Gamma\big(-\frac{1}{4}\big)\Gamma(1)}{\Gamma\big(\frac{3}{4}\big)}\bigg)$$ Using the main property of the Gamma-function $$\Gamma\Big(1-\frac{1}{4}\Big)=-\frac{1}{4}\Gamma\Big(-\frac{1}{4}\Big)\,\Rightarrow\,\Gamma\Big(-\frac{1}{4}\Big)=-4\,\Gamma\Big(\frac{3}{4}\Big)$$ and the relation $$\Gamma\big(\frac{3}{4}\big)\Gamma\big(\frac{1}{4}\big)=\frac{\pi}{\sin\frac{\pi}{4}}=\sqrt2\,\pi$$ $$I=\frac{1}{2}\bigg(4-4\frac{\Gamma\big(\frac{3}{4}\big)\sqrt\pi}{\Gamma\big(\frac{1}{4}\big)}\bigg)=2\bigg(1-\frac{\Gamma^2\big(\frac{3}{4}\big)}{\sqrt{2\pi}}\bigg)$$