Let's assume $a_{1}<a_{2}<\cdots<a_{n}$, then for $1 \leqslant k \leqslant n$, we have
$$\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \geqslant|n+1-2 k|$$

Therefore, $\sum_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3}\right)$
$$\begin{aligned} = & \frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)\left(\frac{3}{4}\left(\left|a_{k}\right|-\left|a_{n+1-k}\right|\right)^{2}+\right. \\ & \left.\frac{1}{4}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{2}\right) \\ \geqslant & \frac{1}{8} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{3} \\ \geqslant & \frac{1}{8} \sum_{k=1}^{n}|n+1-2 k|^{3} \end{aligned}$$

When $n$ is odd, $\sum_{k=1}^{n}|n+1-2 k|^{3}=2 \cdot 2^{3} \cdot \sum_{i=1}^{\frac{n-1}{2}} i^{3}=\frac{1}{4}\left(n^{2}-1\right)^{2}$;
When $n$ is even, $\sum_{k=1}^{n}|n+1-2 k|^{3}=2 \sum_{i=1}^{\frac{n}{2}}(2 i-1)^{3}=2\left(\sum_{j=1}^{n} j^{3}-\right.$
$$\left.\sum_{i=1}^{\frac{n}{2}}(2 i)^{3}\right)=\frac{1}{4} n^{2}\left(n^{2}-2\right)$$

Therefore, when $n$ is odd, $\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32}\left(n^{2}-1\right)^{2}$;
When $n$ is even, $\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32} n^{2}\left(n^{2}-2\right)$, equality holds when $a_{i}=i-\frac{n+1}{2}, i=$ $1,2, \cdots, n$.

Thus, the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$ is $\frac{1}{32}\left(n^{2}-1\right)^{2}$ (for odd $n$), or $\frac{1}{32} n^{2}\left(n^{2}-2\right)$ (for even $n$).