How much time does it typically take for a new mutation to fix in an animal population?
The answer can really take any number. It will all depend upon the size of the population and the selection coefficient on the mutation. If we assume neutral mutations, then we can use Kimura and Ohta (1968)'s result. They showed that the expected time for a neutral allele to reach fixation (given that it will reach fixation) is
$$\bar t(p_0)=-4N\left(\frac{1-p_0}{p_0}\right)\ln(1-p_0),$$
where $p_0$ is the initial frequency and $N$ is the population size. Replacing $p_0$ by $\frac{1}{2N}$ and you get
$$\bar t\left(\frac{1}{2N}\right)=-4N\left(\frac{1-\frac{1}{2N}}{\frac{1}{2N}}\right)\ln\left(1-\frac{1}{2N}\right)$$
which can be simplified
$$\bar t\left(\frac{1}{2N}\right)=-4N\left(2N-1\right)\ln\left(\frac{2N-1}{2N}\right)$$
Stick in any $N$ you want and you'll get your answer. Of course, this equations assume panmixie. In absence of panmixia, then it all depends upon the interconnectedness of the different populations. If some population are perfectly isolated, then a mutation will never fix in the entire species.
How much time does it typically take for a new mutation to happen and fix in an animal population?
The time to wait for the mutation to fix is typically quite negligible comparing to the time it takes for this mutation to happen. Let $\mu$ be the mutation rate. The number of mutations happening each generation is $2 N \mu$ and assuming they are neutral a fraction $\frac{1}{2N}$ of them will fix. Hence, the expected waiting time is $\frac{1}{2 N \mu \frac{1}{2N}} = \frac{1}{\mu}$. In humans, the average $\mu$ is of the order of $10^{-8}$ per site but there is a lot of variation among sequences.
So, if your mutational target is only a single site, then the expected time to fixation of a currently non-existing mutation is $10^8$ generations! If you are considering the entire genome as you mutational target is the entire genome, your mutation rate is of the order of $10$ and the expected number of generations is only $\frac{1}{10}$.