The question doesn't make a whole lot of sense. It is possible to make up examples where the answer is any integer from 0 to n-1.
T1, T2 are non-zero linear functionals on V.
by the rank-nullity theorem: dim(V) = rank(T1) + dim(ker(T1)) = rank(T2) + dim(ker(T2)).
since rank(T1) = rank(T2) = dim(F) = 1, dim(ker(T1)) = dim(ker(T2)) = dim(V) - 1 = n - 1.
so the kernels (nullspaces) in question are two distinct subspaces of dimension n-1 of V.
now, suppose B = {b1,b2,....b(n-1)} is a basis for N1, and that C = {c1,c2,...,c(n-1)} is a basis for N2.
since N1 and N2 are unequal, some element of C is not in span(B), say cj.
then {b1,b2,...,b(n-1),cj} is linearly independent, and thus a basis for V. this shows we can write any element
of V as a linear combination of the b's plus a multiple of cj, that is, as a sum of an element in N1, and one in N2: V = N1 + N2.
since dim(N1+N2) = dim(N1) + dim(N2) - dim(N1∩N2), we have:
n = n-1 + n-1 - dim(N1∩N2), so
dim(N1∩N2) = 2n - 2 - n = n - 2.