There are $\binom{n}{i}$ ways to choose $i$ common elements of two subsets $A,B$. Let $C=A \setminus (A \cap B), D= B \setminus (A \cap B)$ then $|C \cap D|=0$ and $|C \cup D| \le n-i$. There are $\binom{n-i}{l}2^{n-i-l}$ ways to choose pairs of $(C,D)$ so $|C|=l \ge 1$. For $l=0$ then it is $\binom{n-i}{0}2^{n-i}-1$ (minus $1$ to the case $|C|=|D|=0$). Thus, number of pairs $(C,D)$ is $$\frac 12 \left( \sum_{l=0}^{n-i}\binom{n-i}{l}2^{n-i-l}-1 \right)= \frac 12(3^{n-i}-1).$$(An easier way to count number of $(C,D)$ is that each element in the left $n-i$ elements can either belongs to $C$ or $D$ or none of them. Thus, there are $3$ ways to choose each element so there are $3^{n-i}$ ways to choose such pair $(C,D)$ in order including case $|C|=|D|=0$. We take out $1$ from the case $|C|=|D|=0$ and divide by $2$ since we don't care about order of $(C,D)$). The desired sum is \begin{align*} S & =\sum_{i=1}^n i \binom{n}{i} \cdot \frac{1}{2}(3^{n-i}-1), \\ & = \frac n2 \sum_{i=1}^n \binom{n-1}{i-1} (3^{n-i}-1), \\ & = \frac n2 \left( 4^{n-1}- 2^{n-1} \right)=n(2^{2n-3}-2^{n-2}). \end{align*}

For $i \ge 1$, there are $\binom{n}{i} \binom{2^{n-i}}{2}$ pairs of subsets $(A,B)$ so $|A \cap B|=i$. Hence, the desired sum is $S= \sum_{i=1}^{n} i \binom{n}{i}\binom{2^{n-i}}{2}$. We have \begin{align*} S & = n \sum_{i=1}^n \binom{n-1}{i-1} \cdot 2^{n-i-1}(2^{n-i}-1), \\ & = n \left[ 2^{2n-3} \sum_{i=1}^n \binom{n-1}{i-1} \left(\frac{1}{4} \right)^{i-1} - 2^{n-2} \sum_{i=1}^n \binom{n-1}{i-1}\left( \frac 12 \right)^{i-1} \right], \\ & = n \left[ 2^{2n-3} \left( \frac 54 \right)^{n-1}-2^{n-2} \left( \frac 32 \right)^{n-1} \right], \\ & = \frac 12n(5^{n-1}-3^{n-1}). \end{align*}

What I understood from the problem statement is say let $S$ be the set of all subsets $A_{i} \in A$. Find $$\sum\limits_{1 \leq i \neq j \leq 2^{n}} |A_{i} \cap A_{j}| - \sum\limits_{i=1}^{2^{n}} |A_{i}|$$Notice that $|S|=2^{n}$ and every element $x \in A$ is part of $2^{n-1}$ subsets. Then, every element is counted exactly $ 2^{n-1}\choose{2} $ times so: $$\sum\limits_{1 \leq i \neq j \leq 2^{n}} |A_{i} \cap A_{j}| - \sum\limits_{i=1}^{2^{n}} |A_{i}| = n{2^{n-1}\choose{2}}$$