The assertion can be rewritten as
$$\sum_{k=1}^{n} \sum_{j=1}^{k} a_j \le \sum_{k=1}^{n} k^3.$$
It therefore suffices to prove
$$\sum_{j=1}^{k} a_j \le k^3 \quad (6)$$
for every $k = 1, \dots, n$.
In order to prove (6) we fix $k$ and set
$$x_i = \frac{1}{k} - \frac{i}{N} \quad \text{for } i = 1, \dots, k$$
and
$$x_i = \frac{n+1-i}{N^2} \quad \text{for } i = k+1, \dots, n$$
for some integer $N > 0$ to be determined as follows:
The conditions $x_1 > x_2 > \dots > x_n > 0$ are certainly fulfilled in the case $k = n$ and for $k < n$ the only non-trivial relation is $x_k > x_{k+1}$ that is $\frac{1}{k} - \frac{k}{N} > \frac{n-k}{N^2}$. Hence we choose $N > kn$, in order to have $\frac{1}{k} - \frac{k}{N} > \frac{n-k}{N} \ge \frac{n-k}{N^2}$.
The condition $x_1 + \dots + x_n < 1$ means $1 - \frac{k(k+1)}{N} + \frac{(n-k)(n-k+1)}{N^2} < 1$ and will be satisfied for
$$N > \frac{(n-k)(n-k+1)}{k(k+1)}$$
For $N > \max\{kn, \frac{(n-k)(n-k+1)}{k(k+1)}\}$ the numbers $x_1, \dots, x_n$ fulfill the relevant conditions and we conclude
$$\sum_{i=1}^{n} a_i x_i^3 < 1.$$
By taking the limit $N \to \infty$ we get $\sum_{i=1}^{n} a_i \lim_{N \to \infty} x_i^3 \le 1$, which gives $\sum_{i=1}^{k} a_i \frac{1}{k^3} \le 1$, hence the desired estimate (6). $\square$