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 amounts to solving $$ {m+k\choose k}=\frac{\partial}{\partial a_m}\sum_{i=0}^n a_i{i+k\choose k}=\frac{\partial}{\partial a_m}\sum_{i=0}^n a_i^2c_i=2\lambda c_ma_m $$ subject to the quadratic constraint. That is, $a_m=\frac{1}{2\lambda c_m}{m+k\choose k}$ and $$ 1=\sum_{m=0}^n a_m^2c_m=\sum_{m=0}^n \frac{{m+k\choose k}^2}{4\lambda^2 c_m}, $$ which gives $$ \lambda=\frac{1}{2}\sqrt{\sum_{m=0}^n \frac{{m+k\choose k}^2}{c_m}}. $$ Therefore the maximum of $|f(1)|$ is $$ \sum_{m=0}^n{m+k\choose k}a_m=\sum_{m=0}^n \frac{{m+k\choose k}^2}{2\lambda c_m}=\sqrt{\sum_{m=0}^n \frac{{m+k\choose k}^2}{c_m}}=\sqrt{\sum_{m=0}^n \frac{{m+k\choose k}^2(2m+k+1)}{2^{k+1}}}. $$ The terms of the sum are monotonically increasing with $m$, so a loose upper bound is given by $$ |f(1)|\le\sqrt{\frac{n{n+k\choose k}^2(2n+k+1)}{2^{k+1}}}\le\sqrt{\frac{{n+k\choose k}^2(n+k+1)^2}{2^k}}=\frac{k+1}{2^{k/2}}{n+k+1\choose k+1}. $$