The use of hybridisation for transition metal complexes is extremely outdated and often leads to incorrect predictions and rationalisations of transition metal chemistry. If you are using it routinely, please stop using it.
You may wish to first refer to this other answer of mine for a discussion of a very similar case:
Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?
There, the comparison is between a 3d and a 4d metal. However, the factors that affect the geometry of a $\mathrm{d^8}$ complex are the same. They are described in much more detail in that link. To summarise:
Adoption of a square planar geometry is favoured due to occupation of four low-energy orbitals formed.
Adoption of a tetrahedral geometry is favoured due to (a) relief of steric repulsions; and (b) the relief of electron-electron repulsions due to adoption of a high-spin configuration.
When the ligand is $\ce{CN-}$, a strong sigma donor and pi acceptor, the corresponding splitting of the orbitals in $D_\mathrm{4h}$ symmetry will be very large. Consequently, the electronic stabilisation derived upon adoption of a square planar geometry in $\ce{[Ni(CN)4]^2-}$ is much larger than for other complexes, e.g. $\ce{[NiCl4]^2-}$.
It's quite possible that the linear ligand $\ce{CN-}$ also introduces less steric repulsions in a square planar geometry than a halide, like $\ce{Cl-}$ or $\ce{Br-}$, would. However, the main factor is likely to simply be the large electronic stabilisation, which in turn derives from the fact that $\ce{CN-}$ is very high in the spectrochemical series.