As remarked by Iosif Pinelis, this is a matter of law of great numbers; we may also describe it in terms of Bernstein polynomials. Specifically, for $\alpha\ge0$ and $n\ge1$, let $p_n$ be the value of the $n$-th Bernstein polynomial of the function $x^\alpha$ at $1/2$: then $$\sum_{k=0}^nk^\alpha{n\choose k}=p_n 2^nn^\alpha=2^{n-\alpha}n^\alpha(1+o(1)).$$ Moreover, standard facts about convergence give:
For $0\le \alpha\le 1$, since $x^\alpha$ is concave, the sequence $p_n$ is increasing, and it is a modulus of continuity of itself, $$0\le 2^{-\alpha}-p_n \le (4n)^{-\alpha/2}$$
For $ \alpha\ge 1$, since $x^\alpha$ is convex and Lipschitz of constant $\alpha$ on $[0,1]$, the sequence $p_n$ is decreasing, and
$$0\le p_n-2^{-\alpha} \le \frac{\alpha}{2\sqrt n}.$$
Analogous considerations hold for any continuous function in place of $x^\alpha$.