For $0\le k \le n$, we can write $$B_k= \text{Coef. of}~ x^k ~\text{in}~(1+x-x^2)^n$$ $$\implies B_k=\text{Coef. of}~ x^k ~\text{in}~\sum_{j=0}^{n} {n \choose j} x^j(1-x)^j$$ $$=\sum_{j=0}^{n} {n \choose j}\text{Coef. of}~ x^{k-j} ~\text{in}~ \sum_{r=0}^{j} (-1)^r {j \choose r} x^r$$ $$=\sum_{j=0}^{n} {n \choose j}\text{Coef. of}~ x^{k-j} ~\text{in}~ \sum_{r=0}^{j} (-1)^r {j \choose r} x^r$$ $$=\sum_{j=0}^{k} (-1)^{k-j} {n \choose j}{j \choose k-j}........(1) $$ But $f(x)=(1+x-x^2)^n=\sum_{k=0}^{2n} B_k x^k,$ hence the question is: What are $B_k$, when $n+1\le k \le 2n.$ Interestingly, numerical experiments strongly indicate that $|B_k|=|B_{2n-k}|$, so the question becomes: What is the real phase factor in $B_{2n-k}=(-1)^{p(n,k)}B_k$, with proof?
For instance, in the case of $g(x)=(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$, we can prove $A_{2n-k}=A_k$ and $A_k$ is known for $ 0\le k \le n.$
I claim that $B_{2n-k} = (-1)^{k-n} B_k$.
Indeed, $$\sum_{k=0}^{2n} (-1)^{k-n} B_k x^k = (-1)^n \sum_{k=0}^{2n} B_k (-x)^k = (-1)^n (1-x - x^2)^n = (-1 + x + x^2)^n$$ while $$ \sum_{k=0}^{2n} B_{2n-k} x^k = \sum_{k=0}^{2n} B_k x^{2n-k} = x^{2n} \sum_{k=0}^{2n} B_k (1/x)^{k} = x^{2n} (1 + 1/x - 1/x^2)^n = (x^2 + x - 1)^n$$