Is there a closed-form expression for this nested sum? Is there a closed-form expression for this nested sum?
$$s(n)=\sum_{i=1}^n\;\; \sum_{j=i+1}^n \sum_{k=i+j-1}^n1$$
If yes, what is it and how can it be derived?

The usual convention with the sum sign $\sum$  is that $\sum_{k=p}^q a_k:=0$ if $q<p$. This has the following effect: Given an $i$ with $1\leq i\leq n$ the next index variable $j$ has to satisfy
$$i+1\leq j\leq\min\{n, n+1-i\}=n+1-i\ .$$
This then implies that the variable $i$ in fact has to satisfy $2i\leq n$, or
$$1\leq i\leq\left\lfloor{n\over2}\right\rfloor\ .$$
Given $i$ and $j$ with these constraints the innermost sum becomes
$$\sum_{k=i+j-1}^n 1=n+2-i-j\ .$$
The next sum (over $j$) then becomes
$$A_i:=\sum_{j=i+1}^{n+1-i}(n+2-i-j)$$
and has $n+1-2i$ terms. It follows that
$$A_i=(n+1-2i)\>{1\over2}\>(n+2-2i)\qquad\bigl(1\leq i\leq\lfloor n/2\rfloor\bigr)\ .$$
Here we have used that the sum of a finite arithmetic series is the number of terms times the arithmetic mean of its outermost terms. From now on we have to distinguish the cases of even and odd $n$. 
If $n=2m$ then Mathematica produces
$$s(n)=\sum_{i=1}^m A_i={4m^3+3m^2-m\over6}={n(2n^2+3n-2)\over 24}\ .$$
If $n=2m+1$ we similarly obtain
$$s(n)=\sum_{i=1}^m A_i={4m^3+9m^2+5m\over6}={2n^3+3n^2-2n-3\over 24}\ .$$
