Let us check the geometry and magnetic nature of the given complexes:

1. $[Ni(CO)_4]$: In this complex, Ni is in the 0 oxidation state with the electronic configuration $[Ar] 3d^8 4s^2$. Since CO is a strong field ligand, it forces the $4s$ electrons to pair up in the $3d$ orbitals, resulting in a $3d^{10}$ configuration. The empty one $4s$ and three $4p$ orbitals undergo $sp^3$ hybridization, giving a tetrahedral geometry. Since all electrons are paired, it is diamagnetic. Thus, the statement '$[Ni(CO)_4]$, Tetrahedral, paramagnetic' is incorrect.
2. $[Ni(CN)_4]^{2-}$: Ni is in the +2 oxidation state with the configuration $[Ar] 3d^8$. The strong field $CN^-$ ligand pairs up the $3d$ electrons, leaving one empty $3d$ orbital. The complex undergoes $dsp^2$ hybridization, forming a square planar geometry. Due to the absence of unpaired electrons, it is diamagnetic.
3. $[NiCl_4]^{2-}$: Ni is in the +2 oxidation state ($3d^8$). The weak field $Cl^-$ ligand cannot cause pairing of the $3d$ electrons. The complex undergoes $sp^3$ hybridization, resulting in a tetrahedral geometry. The presence of two unpaired $3d$ electrons makes it paramagnetic.