<s>[INST] Dealing with a difficult sum of binomial coefficients, $\sum_{l=0}^{n}\binom{n}{l}^{2}\sum_{j=0}^{2l-n}\binom{l}{j} $ 
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.
 [/INST] By way  of an intermittent  progress report I submit  several integral
representations of the sum
$$\sum_{k=0}^n {n\choose k}^2 \times
\sum_{j=0}^{2k-n} {k\choose j}.$$
They were obtained using the Egorychev method.
The reader  is cordially invited  to comment, verify and  prove these.
Saddle point asymptotics  could be attempted if  these were univariate
integrals.   I present  them in  the  hope that  perhaps a  univariate
generating function can  be found where the catalog  of asymptotics of
coefficient  extractor integrals  could  be  applied. Sometimes  there
exist   substitutions    which   produce   radical    cancellation   /
simplification in these binomial sum integrals.

First representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{1-w} \frac{1}{w^{n+1}}
(1 + w + w^2 z)^n
\; dw\; dz.$$

Second representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} (1+2z)^n (1+z)^n
\; dz
\\ + 
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{2w-1} \frac{1}{(1-w)^{n+1}}
\frac{1}{w^n}
((1-w)^2 z + w)^n
\; dw\; dz.$$

Third representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}}\frac{(1+w)^n}{1-w}
\frac{1}{z-w/(1-w)}
((1+z)^2 w^2 + z)^n
\; dw\; dz.$$

Fourth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}}
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}}
(1 + z + wz)^n
(1 + w^2 z)^n
\; dz\; dw.$$

Fifth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
(1 + w - wz)^n
(z + w^2 - w^2 z)^n
\; dz\; dw $$

Sixth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}} 
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(w+1-z)^n
\; dz\; dw 
\\ - \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{2n}} 
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(w+1-z)^n (1-z + w^2 z)^n
\; dz\; dw .$$
</s>