From the standpoint of intuitionistic logic, multiplicity of roots of generic polynomial is uncomputable due to the inability to compare two real numbers. Even though the roots themselves are computable.
A real number in this context it a regular Cauchy sequence:
$$ x: \mathbb{N} \rightarrow \mathbb{Q}, |x(n) - x(m)| \leq \frac{1}{n} + \frac{1}{m} $$
What if the coefficients of the polymonial are rational numbers?
I mean, if $p(x) = \sum_{i=1}^{n}a_i x^i$, then all the roots (including the complex ones) $x_i, i=1, \ldots n$ are computable. Is it then decidable that $x_i = x_j \lor x_i \neq x_j, i \neq j = 1, \ldots n$?
The presence of multiple roots is detected by the Discriminant which is expressible in terms of the coefficient of the polynomial. So if the coefficients are rational, the presence of multiple roots is decidable.
Edit: I didn't get the OP's question properly in the first place. Concerning the computation of multiplicities once a complete list of zeros is given, it suffices to note that the set $\overline{{\mathbb Q}}\subset{\mathbb C}$ of algebraic numbers is (constructively) a field and has decidable equality. You can find this for example in the introductory chapter of Foundations of Constructive Mathematics by Michael J. Beeson.