How does Ito's formula work at a point where the function is undefined? Consider $(0,...,0)=x\in \mathbb R^{n}$ and $f(x)=\lvert x\rvert^{p},\; p\geq 1$ where $\lvert x\rvert^{p}:=(\sum\limits_{i=1}^{n}x_{i}^{2})^{\frac{p}{2}}$
Now for the Ito vector process $X_{t}:=(X_{t}^{1},...,X_{t}^{n})$ we obtain, where $X_{t}\neq 0$,
$d\lvert X_{t}\rvert^{p}=\lvert X_{t} \rvert^{p-2}\sum\limits_{j=1}^{n}X_{t}^{j}+\frac{1}{2}\sum\limits_{k,l=1}^{n} (p-2)\lvert X_{t}\rvert ^{p-4}X_{t}^{k}X_{t}^{l}+1_{\{k=l\}}\lvert X_{t}\rvert^{p-2}$
I hope that is is correct, but anyhow do I just "ignore" the case when $X_{t}=(0,...,0)$ or is there a special way I can treat it?
Both comments you got are correct, but I'll just expand a bit. Probably you need to handle these on a case-by-case basis. As Michal says, you can add an $\epsilon$ to make the function smooth, and then try to let $\epsilon$ go to zero and show that everything works properly. In many cases in higher dimensions (e.g. Brownian motion in three dimensions or more, as user6... says) the process will avoid the trouble set with probability 1 anyway, since it is transient, and so you would hopefully be ok.
There's a special case worth mentioning, that user6... alludes to. That is when you are working in one dimension, let's just say with Brownian motion, and $f(x) = |x|$. Then, formally, the first derivative is $sgn(x)$, which is 1 for positive $x$ and $-1$ for negative $x$ (its definition at $x=0$ doesn't matter here), and the second derivative is $\delta(x)$, the Dirac delta function. Ito's formula should therefore give
$$ |B_t|-|B_0| = \int_0^t sgn(B_s)dB_s + \int_0^t \delta(B_s)ds. $$
Remarkably, although this is just a formal identity it turns out to be correct, with the last term equal to a new process, the local time of the Brownian motion. This is not a simple matter, and many researchers have worked on developing the properties of this and related processes. A good book if you're interested in this topic is "Markov Processes, Gaussian Processes, and Local Times", by Marcus and Rosen.
Hope that helps.