Let $S$ be the set of lattice points in the rectangle $[1, (n-2)(n-1)] \times [1, n-1]$. Since $n$ is square-free, no point of $S$ lies on the curve $y = \sqrt[3]{nx}$, so the latter splits $S$ into two non-empty sets. The required sum counts the points of $S$ below the curve, while the sum $\sum_{k=1}^{n-1} \lfloor k^3/n \rfloor$ counts the points of $S$ above the curve. Since $|S| = (n-2)(n-1)^2$, it is sufficient to evaluate the latter sum.

To this end, let $t_k = k^3/n - \lfloor k^3/n \rfloor$, $k = 1, 2, \dots, n-1$. Clearly, $0 \le t_k < 1$ for all $k$, and since $n$ is square-free, the $t_k$ are, in fact, all positive. Since $k^3/n + (n-k)^3/n = 3k^2 - 3kn + n^2$ is integral, so is $t_k + t_{n-k}$, and since $0 < t_k + t_{n-k} < 2$, it follows that $t_k + t_{n-k} = 1$. Hence
$$\begin{aligned} n-1 &= \sum_{k=1}^{n-1} (t_k + t_{n-k}) = 2 \sum_{k=1}^{n-1} t_k = 2 \sum_{k=1}^{n-1} \frac{k^3}{n} - 2 \sum_{k=1}^{n-1} \lfloor \frac{k^3}{n} \rfloor \\ &= \frac{n(n-1)^2}{2} - 2 \sum_{k=1}^{n-1} \lfloor \frac{k^3}{n} \rfloor. \end{aligned}$$
Consequently, $\sum_{k=1}^{n-1} \lfloor k^3/n \rfloor = (n-2)(n^2-1)/4$, and
$$\sum_{k=1}^{(n-2)(n-1)} \lfloor \sqrt[3]{kn} \rfloor = (n-2)(n-1)^2 - (n-2)(n^2-1)/4 = (n-2)(n-1)(3n-5)/4.$$