My answer over at cstheory.stackexchange.com involved the expression $$\lim_{n\to \infty} n \left(\frac{2n \choose n}{4^n}\right)^2$$ According to Wolfram Alpha, this expression is at most $\frac{1}{\pi}$.
I would like to know why this is the case, but I don't know enough about these kinds of expressions to figure this out myself.
Using Stirling's formula $${2n \choose n}\approx\frac{4^n}{\sqrt{n\pi}}$$ So
$$L=\lim_{n\to \infty} n \left(\frac{2n \choose n}{4^n}\right)^2\approx\lim_{n\to \infty}\frac{1}{\pi}= \frac{1}{\pi}$$
An alternative to Stirling approximation, if you are familiar with probability, is recall that $$\frac{1}{2^M}{M \choose k}$$
corresponds to a Binomial $(M,p=1/2)$ distribution (probability of getting $k$ heads in $M$ tosses of a fair coin), which, being the sum of $M$ iid Bernoulli variables, has mean $M/2$ and variance $M/4$; and, for large, $M$, we can approximate it by a normal (gaussian) distribution. So, in our case, taking $M=2n$ and $k=n$
$$\frac{1}{2^{2n}}{2n \choose n} \approx \frac{1}{\sqrt{\pi n}}$$
One can also obtain more terms (as an expansion in $n^{-1}$) using Edgeworth expansions.
Use Stirling's formula: $$ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n. $$ Using this formula, we get that $$ \binom{2n}{n} \sim \frac{\sqrt{4\pi n} (2n/e)^{2n}}{2\pi n (n/e)^{2n}} = \frac{4^n}{\sqrt{\pi n}}, $$ and so $$ n \left( \frac{\binom{2n}{n}}{4^n} \right)^2 \sim \frac{1}{\pi}. $$
There are variants of Stirling's formula with bounds on the error terms, for example $$ n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\Theta(1/n)} $$ (Here $\Theta(1/n)$ is some quantity in the range $[-C/n,C/n]$ for some absolute constant $C > 0$.) Using this, we get $$ \binom{2n}{n} = \frac{\sqrt{4\pi n} (2n/e)^{2n} e^{\Theta(1/n)}}{2\pi n (n/e)^{2n} e^{\Theta(1/n)}} = \frac{4^n}{\sqrt{\pi n}} e^{\Theta(1/n)}, $$ and so $$ n \left( \frac{\binom{2n}{n}}{4^n} \right)^2 = \frac{e^{\Theta(1/n)}}{\pi} = \frac{1}{\pi} + \Theta\left(\frac{1}{n}\right). $$ For even more terms, check the Wikipedia article.
Your experiments seem to imply that the sequence $f(n) = n (\binom{2n}{n}/4^n)^2$ is increasing. To check this, let's compute the ratio of two adjacent terms: $$ \begin{align*} \frac{f(n+1)}{f(n)} &= \frac{n+1}{n} \frac{(2n+2)!^2 n!^4}{(2n)!^2 (n+1)!^4} \frac{4^{2n}}{4^{2n+2}} \\ &= \frac{n+1}{n} \frac{(2n+2)^2(2n+1)^2}{(n+1)^4} \frac{1}{16} \\ &= \frac{(2n+1)^2}{4n(n+1)} = \frac{4n^2 + 4n + 1}{4n^2 + 4n} > 1. \end{align*} $$ This shows that $f(n) < 1/\pi$ and $f(n) \to 1/\pi$. Since $f(1) = 1/4$, we get the following explicit formula: $$ f(n) = \frac{1}{4} \prod_{k=1}^{n-1} \frac{4k^2+4k+1}{4k^2+4k}. $$ This also shows that $$ \pi = 4 \prod_{k=1}^\infty \frac{4k^2+4k}{4k^2+4k+1}. $$ Therefore $$ \begin{align*} f(n) &= \frac{1}{\pi} \prod_{k=n}^\infty \frac{4k^2+4k}{4k^2+4k+1} \\ &= \frac{1}{\pi} \prod_{k=n}^\infty \left(1 - \Theta\left(\frac{1}{k^2}\right)\right) \\ &= \frac{1}{\pi} \left(1 - \sum_{k=n}^\infty \Theta\left(\frac{1}{k^2}\right)\right) \\ &= \frac{1}{\pi} \left(1 - \Theta\left(\frac{1}{n}\right)\right) = \frac{1}{\pi} - \Theta\left(\frac{1}{n}\right), \end{align*} $$ as we have obtained before. (Here $\Theta(1/n)$ is some quantity in the range $[C_0/n,C_1/n]$ for some positive constants $0 < C_0 < C_1$.) If we sweat more, we can get better estimates on the error from this formula for $f(n)$; this is left to the reader.
