I am not entirely sure whether I got the point right, so could you please check, whether I wrote the following correctly?
$$\sum_{i=1}^{n}\sum_{j=1}^{n}\binom{\binom{n}{i}}{j}$$
The result should be for $n=2$ this
$$\binom{\binom{2}{1}}{1} + \binom{\binom{2}{1}}{2} + \binom{\binom{2}{2}}{1} + \binom{\binom{2}{2}}{2}$$
The double-sum is equivalent to $$\underbrace{\left[\binom{\binom{n}{1}}{1}+\binom{\binom{n}{1}}{2}+\ldots+\binom{\binom{n}{1}}{n}\right]}_{i=1}+\underbrace{\left[\ldots\right]}_{i=2}+\ldots+\underbrace{\left[\binom{\binom{n}{n}}{1}+\binom{\binom{n}{n}}{2}+\ldots+\binom{\binom{n}{n}}{n}\right]}_{i=n}$$
So for $n=2$, it would have been
$$\underbrace{\left[\binom{\binom{2}{1}}{1}+\binom{\binom{2}{1}}{2}\right]}_{i=1}+\underbrace{\left[\binom{\binom{2}{2}}{1}+\binom{\binom{2}{2}}{2}\right]}_{i=2}=4$$