>! 15 or 18 both seem possible

>! None of their ages can be a prime number or power of prime, since then every other daughter must have a multiple of that prime, and so the sum will be divisible by it.

>! Thus each age must the product of two different primes. The 20 year limit means that these primes must be 2, 3, and 5. At least one of these factors must be absent from one of the ages, so a 3 x 5 = 15 age must exist (and so 7 would be too big). The set of possible ages are 6, 10, 12, 15, and 18, meaning the eldest might be 15 or 18.

>! From here, we must use the knowledge that the sum uniquely determines who is the eldest. The largest possible valid sum is 10, 15, 15, 15, 18, yielding 73. It is clearly impossible to achieve this without the 18. But alas, a unique sum can also be found with 15 as the eldest. 6, 6, 10, 10, 15 yields a sum of 47. I'll check with a computer when the opportunity arises, but this also seems unique.