$\mu=1+\epsilon$ where $\epsilon>0$ holds.



> 
> 1.Is there a good bound for $$T=\frac{\sum\_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}^2}{2^n}?$$
> 
> 
> 


This quantity can be interpreted as $$\sum\_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}\mathbb P(\frac n2+i)$$
where $\mathbb P(\frac n2+i)$ is under bionmial distribution and thus has probability $\frac{\binom{n}{\frac n2 +i}}{2^n}$ which is 'trucated expected value of $\binom{n}{\frac n2 +i}$'.


Computing few values suggests $\log\_2T< n-\log\_2\sqrt{\mu n\ln n}$ at $\mu\rightarrow1^+$.


For example at $n=1000$ to $50000$ gives such margin.



> 
> 2. How large can $\mu$ be for this $\log\_2T< n-\log\_2\sqrt{\mu n\ln n}$ bound to hold up?
> 
> 
> 


Naively I can get $$<\frac{2^n}{\sqrt{n\pi/2}}(1-o(1))$$ by using $$2^nT<\binom{n}{\frac n2}\sum\_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}.$$


I also know we can prove $$\binom{n}{\frac n2 +i}\asymp\frac{2^{nH(\frac12+\frac in)}}{\sqrt{n\pi/2}}$$ approximation.



Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution
Your main conjecture is not quite correct. 


Indeed, for each natural $j$, let $B\_j$ be a random variable (r.v.) with the binomial distribution with parameters $j$ and $1/2$, and let $C\_j$ be an independent copy of $B\_j$. Let also $u:=\sqrt{\mu n\ln n}$. Then for all even natural $n$
$$T/2^n=U\_n:=\sum\_{k\colon\,|k-n/2|<u}P(B\_n=k,C\_n=n-k)
=\sum\_{k=0}^nP(B\_n=k,C\_n=n-k)-R\_n=P(B\_{2n}=n)-R\_n,
$$
where 
$$R\_n:=\sum\_{k\colon\, n/2\ge|k-n/2|\ge u}P(B\_n=k,C\_n=n-k)\le P(B\_n\ge n/2+u)^2
\le e^{-4u^2/n}=1/n^{4\mu^2},
$$
where, in turn, the latter inequality is an instance of an exponential Hoeffding inequality. On the other hand, by Stirling's formula, 
$$P(B\_{2n}=n)\sim1/\sqrt{\pi n}$$
(the asymptotics everywhere here are as $n\to\infty$). So, 
$$T\_n\sim2^n/\sqrt{\pi n}$$
and hence 
$$\log\_2T=n-\log\_2\sqrt{(1+o(1))\pi n}.$$


So, the inequality $\log\_2T<n-\log\_2\sqrt{\mu n\ln n}$ does not hold for any real $\mu\ge1$ if $n$ is large enough. 


