It is a well-known identity that $$\binom{n}{0}^2+\binom{n}{1}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}.$$
By symmetry of the binomial coefficients, this means the ratio $$\dfrac{\binom{2n}{n}}{\binom{n}{0}^2+\binom{n}{1}^2+\cdots+\binom{n}{n/2}^2}$$ is approximately $2$.
What about if we take only the first half of the term in the denominator? How fast does the ratio $$\dfrac{\binom{2n}{n}}{\binom{n}{0}^2+\binom{n}{1}^2+\cdots+\binom{n}{n/4}^2}$$ grow, as $n\rightarrow\infty$?
In the sum $${\binom{n}{n/4}^2+\binom{n}{n/4-1}^2+\cdots+\binom{n}{0}^2}$$ the ratio of one term to the next is initially approximately $9$, and increases thereafter. However, it stays between $9$ and $9 + \varepsilon$ for an arbitrarily long time as $n$ gets larger.
Thus the expression is equivalent to $\binom{n}{n/4}^2(1 + 1/9 + 1/81 + \dots) = (9/8)\binom{n}{n/4}^2$ for large $n$.
By Stirling's formula then, the ratio in your problem is equivalent to $\frac{1}{3}\sqrt{\pi n} \left(\frac{3\sqrt{3}}{4}\right)^n$.