Finding sum : ${\mathop{\sum\sum\sum\sum}_{0\le i\lt j\lt k\lt l\le n }} \,1$

This is just counting the $4$-element subsets of $\{0,1,,\ldots,n\}$. There are $\binom{n+1}4$ of them.

If we use the formula $$ \sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}\tag{1} $$ the method in the question is not all that cumbersome. Each sum building on the previous one, using $(1)$, we start with the sum for $1$ index: $$ \sum_{k=0}^n\binom{k}{0}=\binom{n+1}{1}\tag{2} $$ The sum for $2$ indices is $$ \sum_{k=0}^{n-1}\binom{k+1}{1}=\binom{n+1}{2}\tag{3} $$ The sum for $3$ indices is $$ \sum_{k=0}^{n-1}\binom{k+1}{2}=\binom{n+1}{3}\tag{4} $$ Finally, the sum for $4$ indices is $$ \sum_{k=0}^{n-1}\binom{k+1}{3}=\binom{n+1}{4}\tag{5} $$