What is the oxidation number of the metal in Ni(CO)_4?

Well, we should review the definition of $\text{oxidation state}$, $\text{oxidation number...........}$
And when we do this for $N i {\left(C O\right)}_{4}$, we break the $N i - C$ bonds to get $N {i}^{0}$ and $4 \times \stackrel{-}{:} C \equiv {O}^{+}$, and we are left with a neutral nickel atom, and a neutral carbon monoxide ligand (we normally write the Lewis structure of $C O$ with charge separation, even tho it is a neutral, gaseous molecule). And thus the metal centre has a FORMAL oxidation state of $0$.