Confirming Sandor's answer, I would say that the bound $n+1 \choose 2$ is the best answer in many cases. The ring may be only Noetherian local. Infact there exists always a surjective map from symmetric power to the ordinary power $Sym^2(I)\to I^2$. Tensoring this into $R/\frak{m}$ we see that num of gens of $I^2$ is at most num gens of $Sym^2(I/{\frak{m}} I)$ the latter is degree $2$ monomials in the polynomial ring with $\mu(I)=n$ variables, so that we have $\mu(I^2)\leq {n+1 \choose 2}$(Obviously ;)). The non-obvious fact is that, the equality happens if $Sym^2(I)\to I^2$ is an isomorphism. The class of ideals with this property contains the class of "ideals of linear type" and it is called "syzygetic ideals". Actually $ker(Sym^2(I)\to I^2)=T_2(Spec(R),Spec(R/I))$. For example if an ideal is "Strongly Cohen-Macaulay" and "generically Complete Intersection" then $T_2=0$.
A concrete example is the following: The ideal of the definition of the image of the map $P^1\to P^4$ given by $(s^8:s^5t^3:s^4t^4:s^3t^5:t^8)$.