In particular, if the Lie algebra $\mathfrak{g}$ is nilpotent, is there a tight (tighter than $\frac{n(n+1)}{2}$) upper bound on how many relations are allowed among the basis vectors?
Another question, if a Lie algebra $\mathfrak{g}$ is nilpotent, is it guaranteed that there exists some change of basis such that all of the lie brackets have the form
$[e_i,e_j]=ce_k$
where $e_i,e_j,e_k$ are all basis vectors and $c \in F$? i.e. there are no sums:
$[e_i,e_j]=e_m+e_n$
No, we cannot assume that such a basis exists for a nilpotent Lie algebra. For the title question, "number of allowable Lie brackets" could mean different things. One interpretation is the dimension of $A_n$, the variety of $n$-dimensional Lie algebra laws. Neretin has shown that the bound on this dimension in general is roughly $\frac{2}{27}n^3. $ We have $n^3$ possibilities for defining Lie bracket vectors on an $n$-dimensional vector space, and thus we obtain the trivial bound $n^3/2$ by skew-symmetry. The variety of $n$-dimensional Lie algebra structures on a vector space over $K$ consists of tuples of structure constants in $K^{n^3}$ satisfying skewsymmetry and the Jacobi identity.
The second interpretation is, how many brackets $[e_i,e_j]$ for $i<j$ in the basis vectors we may assume to be zero in an appropriate basis. For example, in dimension 3, we need to have all three brackets in the basis vectors nonzero, as the Lie algebra $\mathfrak{sl}_2$ shows. For the Heisenberg Lie algebra we can find a basis such that only one bracket in basis elements is nonzero. For nilpotent Lie algebras in general, I don't know what the bound is. Certainly there is not always a basis as above.