In the sum $\sum a_n = a_1+a_2+a_3+\cdots +a_n$, the coefficient of $a_i$ is 1
In the sum $\sum \sum a_n = a_1+(a_1+a_2)+(a_1+a_2+a_3)+\cdots +(a_1+a_2+a_3+\cdots+a_n)$, the coefficient of $a_i$ is $n-i+1$
Like wise if we continue for $k$ times, what will be the coefficient of $a_i$ of resultant series.
As zipirovich mentions, the symbol $\Sigma$ is typically meaningless without indices, but you have defined it as an operator:$$ \Sigma a_n = \sum_{i = 1}^n a_i = a_1 + a_2 + ... + a_n $$ This is fine but you ought to be clear you're not using the notation in the standard way. We define $\Sigma^k$ to represent iterating $\Sigma$ $k$ times. Define $\sigma^k_i(n)$ to be the coefficient of $a_i$ in $\Sigma^k a_n$, i.e. $$ \Sigma^k a_n = \sum_{i = 1}^n \sigma_i^k(n) a_i $$ We trivially observe that $\sigma_i^1(n) = 1$. Now observe, $$ \Sigma^k a_n = \Sigma^{k-1}(\Sigma a_n) = \sum_{i=1}^n \sigma_i^{k-1}(n)\Sigma a_i = \sum_{i=1}^n \sigma_i^{k-1}(n)\sum_{j = 1}^i a_j = \sum_{j = 1}^n\left(\sum_{i=j}^n\sigma_i^{k-1}(n)\right)a_j $$ Thus we must have the recursion on $\sigma$: $$ \sigma_j^k(n) = \sum_{i=j}^n\sigma_i^{k-1}(n) $$ For $k = 2$, this yields $\sigma_j^2(n) = \sum_{i=j}^n\sigma_i^{1}(n) = \sum_{i=j}^n 1 = n - j + 1$. For $k = 3$, we must have $$ \sigma_j^3(n) = \sum_{i=j}^n(n - i + 1) = (n - j + 1) + ... + 3 + 2 + 1 = \frac{(n - j + 2)(n - j + 1)}{2} = T_{n - j + 1} $$ where $T_n$ is the $n$th triangular number. In general, it seems that the coefficients will be closely related to simplex numbers, which can be extracted from Pascal's triangle. I believe that it should be straightforward to show that $\sigma_j^k(n)$ is the $(n - j + 1)$th $k$-simplex number.