Here is the value of the maximum, though I'm not sure how it compares to ${n+k+1\choose k+1}$.
Consider the $(k,0)$ Jacobi polynomials $P_n$, which are orthogonal with respect to the weight $(1-x)^k$ on $[-1,1]$. They have squared norm $c_m=\langle P_m,P_m\rangle=\frac{2^{k+1}}{2m+k+1}$ and $P_m(1)={m+k\choose k}$.
Expand $f$ as the sum $\sum_{m=0}^n a_m P_m$. The constraint is that $\sum_{m=0}^n a_m^2c_m=1$, and we would like to maximize $\sum_{m=0}^n a_m{m+k\choose k}$.
We can apply the method of Lagrange multipliers, which involves solving ${m+k\choose k}=2\lambda c_ma_m$ subject to the quadratic constraint. That is, $$ 1=\sum_{m=0}^n a_m^2c_m=\sum_{m=0}^n \frac{{m+k\choose k}^2}{4\lambda^2 c_m}. $$ Therefore the maximum of $|f(1)|$ is $$ \sum_{m=0}^n \frac{{m+k\choose k}^2}{2\lambda c_m}=\sqrt{\sum_{m=0}^n \frac{{m+k\choose k}^2}{c_m}}=\sqrt{\sum_{m=0}^n \frac{{m+k\choose k}^2(2m+k+1)}{2^{k+1}}}. $$