This is what I think the technical statement of CLT is:
If we consider $\overline{X}_{n}$ coming from a sample of $\mathcal n$ independent and identically distributed random variables $X_{i}$ with mean $\mu$ and variance $\sigma^{2}$ then as n $\rightarrow \infty$
$\sqrt n \left( \left(\frac{1}{n} \ \sum_{i=1}^{n} X_{i} \right) - \mu \right) \approx N(0,\sigma^{2})$
Question: I've read textbooks which claim $n \ge 30$ without offering any support for the claim, why $n \ge 30$?
There are quantitative versions of the CLT which provide an estimate of the rate of convergence. One is given at http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem. This says that the sequence of CDFs of the rescaled sample means converges uniformly to the appropriate normal CDF, provided the third moment of the variables exists.
Assuming the summands have finite third absolute moment, we get $|F_n - \Phi| \leq \frac{C \rho}{\sigma^3 \sqrt{n}}$ where $\sigma$ is the standard deviation of the summands and $\rho$ is the absolute third moment of the summands. Here $C$ does not depend on the variables, and has been shown to be less than $\frac{1}{2}$. So the practical convergence rate depends on how big $\rho$ is relative to $\sigma$.
Let's try this with the binomial. The standard deviation is $(p(1-p))^{1/2}$. The third absolute moment is $p (1-p)^3 + (1-p) p^3 = p (1-p) ((1-p)^2+p^2)=p (1-p) (1-2p+2p^2)$. The factor of $(1-2p+2p^2)$ is between $1/2$ and $1$, so we can essentially ignore that. The important thing is that $\frac{\rho}{\sigma^3}$ is on the order of $(p (1-p))^{-1/2}$. So for the binomial, the theorem estimates the error as being at most $\frac{1}{2 p^{1/2} (1-p)^{1/2} \sqrt{n}}$. Making this small could require $n$ to be quite large, if $p$ is close to $0$ or $1$.
A quantitative example: take $p=0.01$, then $\frac{1}{2 p^{1/2} (1-p)^{1/2}}$ is a little larger than $5$. In this case, the theorem only guarantees that the error is less than $0.1$ uniformly if $n>2500$. In reality this is probably rather blunt (indeed, a sharp argument for this special case can be found at http://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem), but it gives an idea of how slow the convergence rate could be in the worst cases.