First, count the total number of ways that the fish can eat the pellets (assume they all get eaten). For this, we construct all multisets of cardinality 10, chosen from a finite set of cardinality 7. (We keeping choosing fish 10 times, with no restriction on how many times ($\leq10$) that we can choose each fish.) The number of such distinct multisets is $C(7+10-1,10)=8008$.
Now follow your idea of counting the number of ways that no fish eats more than 2 pellets. If we write $p_i$ for the number of pellets eaten by fish$_i$, then $p_1+p_2+\cdots+p_7=10,$ with the restriction that $0\leq p_i\leq 2$ for each $i$. We can count these as follows: the key is to consider how many fish get 2 pellets.
Case 1: Five fish eat 2 pellets, two fish eat 0 pellets. This can happen $C(7,2)=21$ ways.
Case 2: Four fish eat 2 pellets, two fish eat 1 pellet, one fish eats 0 pellets. This can happen $7\times C(6,2)=105$ ways.
Case 3: Three fish eat 2 pellets, four fish eat 1 pellet. This can happen $C(7,3)=35$ ways.
These are the only possibilities, so there is a total of 161 ways that no fish gets more than 2 pellets. So the probability that at least one fish gets more that 2 is $\frac{8008-161}{8008}\simeq 0.98$.
Aside: The total number of ways that the fish can eat the pellets is not $7^{10}$. This result could be arrived at as follows: How many ways can the first pellet be eaten? Ans: 7. ... How many ways can the 10th pellet be eaten? Ans: 7. How many ways in total? Ans: $7\times\cdots\times7=7^{10}$.
But this fails to distinguish between these two outcomes: (i) Fish$_1$ eats pellets 1-9, fish$_2$ eats pellet 10; (ii) Fish$_1$ eats pellets 2-10, fish$_2$ eats pellet 1.
But these outcomes are identical in terms of the question, as they both correspond to "fish$_1$ eats nine pellets, fish$_2$ eats one pellet". Thus $7^{10}$ includes lots of multiple counting of the same outcome.
Further aside: This is wrong! 8008 is indeed the number of different possible ways that the pellets can be eaten (assuming distinct fish), but these outcomes are not equally likely. $7^{10}$ is the number of equally likely outcomes.