I'm gonna go with
8
This is the smallest possible big ball weight because
Choosing the weights 1, 2, 3, 5 and 8, every possible combination of exactly three balls yields a unique sum. Therefore, every weighing reveals exactly which three ball were weighed. There is no combination with this property where the heaviest ball is lighter than 8.
1,4,6,7,8 ("differences between the numbers in reverse order", obtainable by subtracting each number in the original sequence from 9) would also work, but I like 1,2,3,5,8 better.
If (and only if) the weight distribution satisfies the above condition, a guaranteed identification strategy exists. Here’s an example strategy: label the unknown balls A, B, C, D, and E, and then weigh the balls like so
A+B+C (weighing X)
A+B+D (weighing Y)
A+D+E (weighing Z)
This will identify each ball:
Rebember that each weighing exactly reveals which three balls were weighed. Then
* A is the ball that is present both in X and Z
* B is the other ball besides A that is present in both X and Y
* C is the ball present in X that isn't A or B
* D is the ball present in Y that isn't A or B
* E is the remaining ball
For example, using the distribution with the lightest possible big ball:
if the shuffle happens to be A=8, B=1, C=2, D=5, E=3, then the measurements are
X: 11 -> 1, 2 and 8
Y: 14 -> 1, 5 and 8
Z: 16 -> 3, 5 and 8
and the balls are
A: present in X and Z -> 8
B: present in X and Y, is not A -> 1
C: present in X, isn't A or B -> 2
D: present in Y, isn't A or B -> 5
E: the remaining ball -> 3