The sum is extremely unimodal, so we want to approximate it using its largest term. Taking logs of each term, we see that the max occurs when $(26-n) \log n$ peaks, and taking derivatives gives $x+x \log x=26$. From here it's easy to see that the answer is around 10, and slightly less (it's actually about 8.3, but in any case it's hard to find powers of anything except 10). Thus the largest term will be something like $10^{16}$, which is already an order of magnitude within the desired answer $6.6 \times 10^{16}$. To do better we'd really need to understand the behavior of the function $x^{26-x}$, but what approximately happens is that only the four or five largest terms in the sum are of any substantial size; thus it is reasonable here to pick some constant from 4 to 20 to multiply our guess $10^{16}$; any guess between $4.0 \times 10^{16}$ and $2.0 \times 10^{17}$ is reasonable.