With $a,b,c \in \mathbb{Z},$ by congruence computation we get that $n= a^5+b^5+c^5$ implies $n \not \equiv 4,5,6,7 \pmod{11} $.
For $a,b,c \in \{-100,-99, \dots , 99, 100\}$, the set of integers $n \in \{0,1,\dots , 99,100 \}$ we get is
$$\{ 0, 1, 2, 3, 12, 30, 31, 32, 33, 34, 63, 64, 65, 96 \}$$
The smallest non-obvious solution being $n=12$,
$$13^5 + 16^5 - 17^5 = 12$$
The point is that it is exactly the same for $a,b,c \in \{-10000,-9999, \dots , 9999, 10000\}$, so that we could expect that there is no other natural number $n \le 100$ representable like that. Nevertheless according to what happens for cubes (see here), we could also expect the existence of such representations with large integers. By the congruences above, the smallest natural numbers to look is $n=8$.
Question: Can a sum of three fifth power of integers be $8$?
Next we should look to $n = 9,10,11,13,14,19,20,21,22,23,24,28,29, \dots$