Does the analytic continuation of $f$ always exist? Let $f(z)$ be a holomorphic funtion on region $\Omega$. Then, does the analytic continuation of $f$ always exist? Note that $f$ is always the analytic continuation of itself, so I exclude this case. 
If the answer is yes, then what is the analytic continuation of $\sum_{n=0}^\infty z^n$, defined on the unit disk?

The analytic continuation does not always exist. The region $\Omega$ could be "maximal" in that sense.
In case of $f(z)=\sum_{n=0}^\infty z^n$, the continuation exists on $\mathbb{C}\setminus \{ 1 \}$ and is $f(z)=\frac{1}{1-z}$ as mentioned in a comment to your question.
