It is possible to reverse the birthday bound calculation. You can get an easily computable approximation using the expected number of collisions:
If you had random $n$-bit salts, after $k$ values you would expect $2^{-n}\binom{k}{2}$ collisions. If the collisions are rare, they are mostly single collisions, so there are approximately $u = k-2^{1-n}\binom{k}{2}$ unique, non-colliding values. So your results $k = 1.8 \cdot 10^7$ and $u \approx 1.7 \cdot 10^7$ could be generated with $n \approx 28.3$. That is, they could be random numbers between zero and 340 million, give or take.
More exact formulas are found in the answers to this question Neil Slater linked in the comments. With your numbers they give an answer of around 27.3. This makes sense, because the approximation above ignored multi-collisions, which leave more unique values than estimated.
However, that does not imply they have 27-28 bits of entropy. They could have considerably less. For example, they could instead be the results of a counter that wraps around after ~17500000 items. There is no way to know with just the data you gave. They very likely do not have more than about 28 bits of entropy each, however.