Let us find the hybridization of the central metal atom in each complex:

1. In $[Ni(CO)_4]$, Ni is in the $0$ oxidation state with the electronic configuration $[Ar] 3d^8 4s^2$. CO is a strong field ligand, which forces the pairing of $4s$ electrons into the $3d$ orbitals, resulting in a $3d^{10}$ configuration. The one $4s$ and three $4p$ orbitals undergo $sp^3$ hybridization, making it a tetrahedral complex.
2. In $[Ni(CN)_4]^{2-}$, Ni is in the $+2$ oxidation state with the configuration $[Ar] 3d^8$. The $CN^-$ ion is a strong field ligand, which causes the pairing of the two unpaired $3d$ electrons, leaving one $3d$ orbital empty. The empty $3d$, one $4s$, and two $4p$ orbitals undergo $dsp^2$ hybridization, resulting in a square planar geometry.
3. In $[NiCl_4]^{2-}$, Ni is in the $+2$ oxidation state with the configuration $[Ar] 3d^8$. The $Cl^-$ ion is a weak field ligand, so it cannot cause the pairing of the $3d$ electrons. The one $4s$ and three $4p$ orbitals undergo $sp^3$ hybridization, leading to a tetrahedral geometry. Therefore, the hybridization states are $sp^3, dsp^2, sp^3$ respectively.