I'm not sure of your math-stat background in this problem, so bear with me.
For a big sample size [tex]\hat p[/tex] has approximately a normal distribution, right? You can approximate the distribution of [tex]\hat{p}^2[/tex] (it will also turn out to be a normal distribution - look in (say) Hogg/Craig or any introductory math stat book for the idea, or write back and I can put the method here), and then you can get an approximate confidence interval for [tex]p^2[/tex]
Note - just so I don't have to post it:
If an estimate [tex]X_n[/tex] for some parameter [tex]\theta[/tex] satisfies
[tex]
\sqrt n \left(X_n - \theta \right) \sim n(0, \sigma^2)[/tex]
(the [tex]\sim[/tex] means "tends to a normal distribution as [tex]n \to \infty[/tex] - i.e., it represents convergence in distribution)
then for a function [tex]f[/tex] that is continuous and has a non-zero derivative at [tex]\theta[/tex] it is true that
[tex]
\sqrt{n} \left(f(X_n) - f(\theta)\right) \sim n(0, \sigma^2 f'(\theta) \right)[/tex]
Your statistic is the sample proportion, the parameter is [tex]p[/tex], and the function is [tex]f(x) = x^2[/tex]