Is there tight estimates for the following logarithmic summation ($\gamma,\gamma'\in(0,1)$ and $\mu,\mu'>0$)


$$\log\_2\Bigg(\sum\_{t=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum\_{\ell=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum\_{k=\frac t2-n^{\gamma'}\sqrt{\mu'\ln n}}^{\frac t2+n^{\gamma'}\sqrt{\mu'\ln n}}\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)?$$


I am hoping it might give $n - f(n)$ bound where $f(n)$ is $\omega(\ln n)$ or at least $\Omega(1)$ for diagonal case of $\gamma=\gamma'=\frac12$ at some $\mu,\mu'>0$.


Relevant problem is in [Tight estimates for binomial summation](https://mathoverflow.net/questions/350336/tight-estimates-for-binomial-summation/) (and perhaps an upper bound **possible** might be $$\log\_2\Bigg(\underbrace{n^{2\gamma}\mu(\ln n)}\_{\substack{\mbox{coming from}\\\mbox{outer two}\\\mbox{summations}}}\binom{n}{n/4}\Bigg)<n H(\frac14+\epsilon)<0.82n$$ at any $\epsilon>0$ since $t/2=n/4$ might be close to the value that yields the bound (together with loose upper bound from [Vandermonde's identity](https://en.wikipedia.org/wiki/Vandermonde%27s_identity) for inner sum)).



Tight sublinear estimates for a triple partial binomial summation
This conjecture does not hold even in the case $\gamma=\gamma'=1/2$. 


Indeed, consider the values of $\ell,t,k$ such that 
$$|\ell-n/2|\ll\sqrt n,\ |t-n/2|\ll\sqrt n,\ |k-t/2|\ll\sqrt n,$$
where $A\ll B$ or, equivalently, $B\gg A$ means that $|A|\le CB$ for some universal real constant $C>0$; as usual, $A\asymp B$ means that $A\ll B\ll A$. 
By what was shown in [this answer](https://mathoverflow.net/a/350722/36721), 
$$\binom\ell k\asymp\frac{2^\ell}{\sqrt\ell}\,e^{-u^2/2},$$
where 
$$u:=\frac{k-\ell/2}{\sqrt{\ell/2}}\ll1, $$
so that 
$$\binom\ell k\asymp\frac{2^\ell}{\sqrt n}.$$
Similarly,
$$\binom{n-\ell}{t-k}\asymp\frac{2^{n-\ell}}{\sqrt n},$$
whence
$$\binom\ell k\binom{n-\ell}{t-k}\asymp\frac{2^n}n.$$
Hence (in the case $\gamma=\gamma'=1/2$), your big triple sum is 
$\gg \dfrac{2^n}n\,n^{3/2}$
and hence 
$$\log\_2(\text{the triple sum})-n\gg\ln n.$$


