See my second (according to date) answer to this question.
Elaborating (based on results from the aforementioned thread, and using the same notation).
Let $\bar Z_n = \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$ be the corresponding quotient approximating $\pi/4$ (with probability $1$, $\bar Z_n \to \pi/4$ as $n \to \infty$). It has expectation and variance given by $ {\rm E}[\bar Z_n ] = \frac{\pi }{4}, \;\; {\rm Var}[\bar Z_n ] = \frac{\pi }{{4n}}\Big(1 - \frac{\pi }{4}\Big) < \frac{{10}}{{59n}}, $ leading to the following probabilistic error bound: $ {\rm P}\bigg[\bigg|\bar Z_n - \frac{\pi }{4} \bigg| \geq \varepsilon \bigg] < \frac{{10}}{{59n\varepsilon ^2 }}. $ Thus, for example, $ {\rm P}\bigg[\bigg|\bar Z_{10^6} - \frac{\pi }{4} \bigg| \geq 10^{-2} \bigg] < \frac{{1}}{{590 }}. $ Further, the central limit theorem leads to the following approximation (in distribution): $ \bar Z_n - \frac{\pi }{4} \approx \frac{\sigma }{{\sqrt n }}\xi, $ where $ \sigma^2 = {\rm Var}[Z_1] = \frac{\pi }{4} \Big(1 - \frac{\pi }{4}\Big) < \frac{10}{59} $ (hence $\sigma \approx 0.41$) and $\xi \sim {\rm Normal}(0,1)$. For example, for $n=10^6$, this gives $ \bar Z_{10^6} - \frac{\pi }{4} \approx 4.1 \times 10^{-4} \xi. $ Now, you can use a Normal Distribution Calculator to get a probabilistic estimate of the error. For example, ${\rm P}(|\xi| \leq 1) \approx 0.682689$, ${\rm P}(|\xi| \leq 2) \approx 0.9545$, and ${\rm P}(|\xi| \leq 3) \approx 0.9973$. Hence, with high probability, the absolute error $|\bar Z_{10^6} - \pi/4|$ will be less than, say, $10^{-3}$.
Finally, it is interesting to compare the above approximation with the following one (see my first answer to the related question). Let $\bar Y_n = \frac{{\sum\nolimits_{i = 1}^n {\sqrt {1 - U_i^2 } } }}{n}$, where $U_i$ are independent uniform$[0,1]$ variables (it too converges, with probability $1$, to $\pi / 4$ as $n \to \infty$). $\bar Y_n$ has expectation and variance given by $ {\rm E}[\bar Y_n ] = \frac{\pi }{4}, \;\; {\rm Var}[\bar Y_n ] = \frac{1}{n} \bigg \{ \frac{2}{3} - \frac {{\pi ^2 }}{{16}}\bigg\} < \frac{1}{{20n}}, $ leading to $ {\rm P}\Big[\Big|\bar Y_n - \frac{\pi }{4} \Big| \geq \varepsilon \Big] < \frac{1}{{20n\varepsilon ^2 }}. $ As before, the central limit theorem leads to the approximation $ \bar Y_n - \frac{\pi }{4} \approx \frac{\sigma }{{\sqrt n }}\xi $ (with $\xi \sim {\rm Normal}(0,1)$), but this time $ \sigma^2 = {\rm Var}[Y_1] = \frac{2}{3} - \frac{{\pi ^2 }}{{16}} < \frac{1}{20} $ (hence $\sigma \approx 0.223$). Note that the variance in this case is considerably smaller than the one in the previous case.