Simplifying sigma of sigma of sigma

Write $$\sum_{i=1}^{z-2}{ (z - 1) - i}+\sum_{j = i+1}^{z-2} {(z - 1) - j} + \sum_{k = j+1}^{z-2} {(z - 1) - k} + \cdots = \sum_{i =1}^{z-2} {(z - 1)} - \sum_{i = 1}^{z-2} {i}+\left(\sum_{j =1}^{z-2} {(z - 1) - j} - \sum_{j =1}^{i} {(z - 1) - j} \right)+ \left(\sum_{k =1}^{z-2}{ (z - 1) - k} -\sum_{k =1}^{j}{ (z - 1) - k}\right) + \cdots,$$ then use the facts that $$\sum_1^n =n, \,\,\sum_1^n i = \frac{n(n+1)}{2}$$ repeatedly.

Let $n=z-1$ to simplify the notation and let's generalize the problem to finding the general sum

$$I_n = \sum_{i_1 = 1}^{n-1}\left(f(n-i_1) + \left ( \sum_{i_2 = i_1+1}^{n-1} f(n-i_2) \ + \ldots + \sum_{i_{n-1}=1+i_{n-2}}^{n-1}f(n-i_{n-1})\right ) \right)$$ In your case we have $f(x) = x$. Note that there will only be $n-1 = z-2$ sums as another sum would be empty (the smallest possible value for $i_k$ is $k$ which is greater than $n-1$ for $i_n$).

By a change of the summation indices $i_k \to n-i_k$ we can write the sum on the form $$I_n = \sum_{i_1 = 1}^{n-1}\left(f(i_1) + \left ( \sum_{i_2 = 1}^{i_1-1} f(i_2) \ + \left ( \sum_{i_3 = 1}^{i_2-1} f(i_3) + \ldots + \sum_{i_{n-1}=1}^{i_{n-2}-1}f(i_{n-1})\right ) \right) \right)$$ which has the nice property than we have removed the $n$ dependence on each sum but the first and which allows us to show that $I_n$ satisfy a simple recurrence relation.

Take the expression for $I_{n+1}$ and split the first sum into that over $i_1=1,2,\ldots,n-1$ and that over $i_1=n$. Notice that for the former case the innermost sum is always empty for the reason we said above so this is just $I_n$ and for the latter case we are just left with $f(n$) plus the same expression as for $I_n$ just with a different name for the summations labels $i_k\to i_{k+1}$. In formulas $$I_{n+1} = I_n + \left(f(n) + \left ( \sum_{i_2 = 1}^{n-1} f(i_2) \ + \left ( \sum_{i_3 = 1}^{i_2-1} f(i_3) + \ldots + \sum_{i_{n}=1}^{i_{n-1}-1}f(i_{n})\right ) \right) \right) \\= I_n + (f(n)+I_n)$$

Thus the complicated looking sum satisfy the simple relation

$$I_{n+1} = 2I_n + f(n)$$

For your special case of $f(x) = x$ the solution is $I_n = 2^n - n - 1 = 2^{z-1} - z$ which is the first Eulerian number (which suggest there likely is some simple combinatorical proof of the identity). As a double check this agrees with a direct numerical computation for $z=3,4,5,\ldots$ which gives $1,4,11,26,\ldots$.

Other simple cases for $f$ are also solveable. For example if $f(n) = 1$ then $I_{n} = 2^{n-1}-1$ and in general if $g$ is a polynomial of degree $n$ then $I_n = c2^n + g(n)$ where $c$ is a constant and $g$ is a polynomial of degree $n$.