TITLE: Dealing with a difficult sum of binomial coefficients, $\sum_{l=0}^{n}\binom{n}{l}^{2}\sum_{j=0}^{2l-n}\binom{l}{j} $
QUESTION [12 upvotes]: I am interested in finding an upper bound for the sum 
  $$F(n)= \sum_{l=0}^{n}\binom{n}{l}^{2}\;\sum_{j=0}^{2l-n}\binom{l}{j}$$ 

Ideally it should be possible to evaluate it exactly using some combinatorial identities or generating functions. Note that the $\binom{n}{l}$ term is squared, and that when $2l-n<0$, we define the inner to be zero.
Simple upper bound: By noticing that $2l-n\leq n$, we have that  $$\sum_{j=0}^{2l-n}\binom{l}{j}\leq 2^n,$$ and so $$F(n)\leq 2^n \sum_{l=0}^{n}\binom{n}{l}^{2}=2^n\binom{2n}{n}.$$ (See this question for a proof that $\sum_{l=0}^{n}\binom{n}{l}^{2}=\binom{2n}{n}$) Modifying the above based on when the inner sum is zero, that is when $2l\leq n$, we can replace this upper bound by $$F(n)\leq 2^{n-1}\binom{2n}{n}.$$ However it should be possible to do significantly better than this.

REPLY [3 votes]: First time contributor to Stack Exchange, so pardon my typesetting. My proof could use some more rigor, so I’ll only sketch the idea.
From an integral representation of the exponential generating function for the numbers F(n), I was able to do classic saddle point analysis.  I then used ‘de-Poissionization’ to extract an asymptotic formula for F(n).  I will give a closed form for only the most important entity, A, since the other factors are an awful combination of derivatives of a complicated function that is evaluated at the saddle point.
$$ F(n) \approx C \frac{A^{n+1}}{n+1} \bigl( 1 - \frac{A\, B}{n+2} \bigr) $$
$$ A=\bigl(\sqrt{y_0}+\sqrt{1+1/y_0}\bigr)^2 \, , y_0=\frac{1}{3}\bigl( -1 + \tau^{1/3} + \tau^{-1/3} \bigr) \, ,
\tau=\frac{1}{2}(25+3\sqrt{69}) .$$
Numerically,
$$ A=5.729031537980930838, B=4.1847320149263052, C=0.250939299689850514 .$$
As an example, F(2000)=1.03227E1513 and the approximation is F*(2000)=1.0318E1513, so that about 4 significant digits are obtained.  The proposer wanted a bound, and since the second term in the asymptotic series is negative, the first term is sufficient, for n large enough.