Can we determine if a filter is butterworth or chebyshev from its physical topology?
As pointed out in a comment by Jason R, you can obviously come up with some random coefficients for the numerator and denominator polynomials of the filter's transfer function, and you'll end up with arbitrary filters that do not satisfy the definitions of any of the filter types you mentioned (even if you make sure that the filter is causal and stable).

Concerning your second question, the answer is also no. A Butterworth and a Chebyshev (type I) filter have the same structure, just the element values are different. The structure of an elliptic filter is different because it has zeros at finite frequencies. For passive filters, these are achieved either by parallel circuits of L and C in the signal path, or by series circuits of L and C towards ground. However, this is also true for type II Chebyshev filters, which also have zeros at finite frequencies. So if your passive filter topology has LC resonators (parallel in the signal path, or serial towards ground) you know that the filter can't have a Butterworth, Bessel, or type I Chebyshev characteristic. It could be an elliptic or a Chebyshev (type II) filter, but it needn't be. If the structure has no LC resonators, the opposite is true: it can't be an elliptic or a Chebyshev (type II) filter, but it could be a Bessel, Butterworth, or Chebyshev (type I) filter. But again, it could also be some non-standard filter. It all depends on the element values.