The problem is equivalent to the one in OEIS A008793, a.k.a. "number of ways to tile hexagon of edge n with diamonds of side 1." Notice that there is a bjiection between such a tiling and the number of ways to stack some unit cubes alongside a corner of an $n \times n \times n$ box (see the Art of Problem Solving logo as an example, also known as 3 -dimensional Young diagrams), where this problem $n=5$. It is known that there are $\binom{2 n}{n}=252$ ways to stack one layer (since each way correspond a way to walk from a corner of a 5 by 5 grid to the opposite one), so $\frac{252^{5}}{5!} \approx 8 \times 10^{9}$ gives a somewhat loose upper bound (generate five layers and sort them by size, and hope that it will be a valid stack in general). This result can be improved by dividing out a reasonable constant factor after considering the probability that sorting by size indeed gives a valid stack (for example, it would be fair to guess that there is about a $\frac{1}{4!}$ chance that the first row of each layer will be in the right order, given that each row has a small role in determining the final size of the layer; dividing 24 from the previous result gives a very close guess). In general, a guess anywhere between $10^{8}$ and $10^{9}$ is a fair guess.