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 since $x^\alpha$ 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, the sequence $p_n$ is decreasing, and since $x^\alpha$ is Lipschitz of constant $\alpha$ on $[0,1]$
$$0\le p_n-2^{-\alpha} \le \frac{\alpha}{2\sqrt n}.$$
Analogous considerations hold for any continuous function in place of $x^\alpha$.
[edit] as to the above bounds on the remainder, the general fact is: Given $f\in C^0([0,1])$, $\omega$ a concave modulus of continuity for $f$, and $x\in[0,1]$, the elementary inequality holds: $$|f(x)-B_nf(x) |\le \omega\Big(\sqrt{\frac{x(1-x)}n}\Big)$$ whence the uniform bound
$$\|f-B_nf\|_\infty\le \omega\Big( \frac1{2\sqrt n}\Big).$$
For $x^\alpha$ we can take $\omega(t):=t^\alpha$ for $0\le\alpha\le1$ and $\omega(t)=\alpha t$ for $\alpha\ge 1$.