The standard approach here would be to take a to be determined sample size $n$.
Under the assumption of the coin being not biased, which should be your null hypothesis, the average times you see tails equals $1/2$. The variance of this average is given by $\frac{1}{2}\cdot\left(1-\frac{1}{2}\right)/n = 1/(4n)$. Based on the central limit theorem you can conclude that (approximately) with $95\%$ certainty
$$\overline{X}_n \in \left[\frac{1}{2} \pm 2\cdot\frac{1}{2}\sqrt{\frac{1}{n}} \right] $$
holds, where I took $2$ as an approximation of $z_{0.975}$. So, if your average is outside of this confidence interval, it might be the case that your coin is biased.
You could formulate a similar result for a general $p$ as probability for the coin being tails.
Problem is that there is no general way of approaching your question because it depends on how rigged the coin may be. For example if the probability $p$ of getting tails equals $1/4$, a relatively small amount of trials is needed to conclude biasedness of the coin. However, if the probability $p$ of getting tails equals $0.49$, way more trials are needed. Suppose this is the case, then you need approximately $10.000 = 1/(0.5 - 0.49)^2$ trials. This amount increases exponentially as $p$ tends to $0.5$.
I hope this helps you a little bit and if you have any question don't hesitate to ask them!