Consider  the set of eight vectors $V = \begin{Bmatrix} a\hat{i} + b\hat{j} + c\hat{k} : a, b, c \,\, \in (-1, 1)\end{Bmatrix}.$ Three non-coplanar vectors can be chosen from V in 4 $2^n$ ways. Then, n is 

5 

Eight vectors V are

$\vec{u} = \hat{i} + \hat{j} + \hat{k} , \vec{v} = \hat{i} + \hat{j} - \hat{k}, \vec{w}= \hat{i} - \hat{j} + \hat{k}, \vec{\alpha }= \hat{i} - \hat{j} - \hat{k}$

$\vec{u'} = \hat{i} - \hat{j} - \hat{k} , \vec{v'} = -\hat{i} - \hat{j} + \hat{k}, \vec{w'}= -\hat{i} - \hat{j} + \hat{k}, \vec{\alpha' }= -\hat{i} + \hat{j} + \hat{k}$

Clearly, $\vec{u}, \vec{u'}; \vec{v}, \vec{v'}; \vec{w}, \vec{w'}$ and $\vec{\alpha}, \vec{\alpha'}$ are four pairs of unlike parallel vectors. If we take any one pair out of these four pairs, and one vector from the remaining 6 vectors, we obtain three coplanar vectors.

∴ Total number of ways of selecting three coplanar vectors $= {^6C}_1 × {^4C}_1 = 24$

Also, 

Number of ways selecting 3 vectors out of given 8 vectors $= {^8C}_3$

∴ Number of ways of selecting 3 non-planar vectors $= {^8C}_3 - 24 = 32 = 2^5$

Hence, $n = 5 $