If I flip a coin 1000 times in a row and it lands on heads all 1000 times, what is the probability that it's an unfair coin? Consider a two-sided coin. If I flip it $1000$ times and it lands heads up for each flip, what is the probability that the coin is unfair, and how do we quantify that if it is unfair? 
Furthermore, would it still be considered unfair for $50$ straight heads? $20$? $7$?

For "almost all" priors, the answer is 100%: you have to specify an interval of "fairness" in order to get a nonzero value. Why? Because the bias $p$ can be any real number between $0$ and $1$, so the probability that it is a very specific real number is $1/\infty = 0$; hence the probability that it is $1/2$ is 0.
However, what I can tell you is that after your tosses, the heads probability is $$\frac{h + 1}{t + h + 1 + 1} = \frac{1000 + 1}{0 + 1000 + 1 + 1} = \frac{1001}{1002}$$
...assuming you initially believed the coin was fair (see @leonbloy's comment under the question).  
More generally, if the heads probability $p$ has prior distribution $\Pr(p)$ then the answer is:
$$\frac{\int_0^1 p\,p^{1000}(1-p)^{0}\Pr(p) \,dp}{\int_0^1 \phantom{p\,} p^{1000}(1-p)^{0}\Pr(p) \,dp}$$
Notice that for the uniform prior $\Pr(p) = 1$, this degenerates to what I gave above.
The derivation is longer and likely to be much more complicated than you might expect, so I'll omit it to save typing... if you really want to see it, let me know.
(Basically, you need to apply Bayes's rule and simplify.)
