= Solution
For the constant <monomial> $f_0=1$, the <binomial theorem> gives $B_n(f_0)(x)=(x+1-x)^n=1$. For $f_1$, the <binomial coefficient> identity $k\binom nk=n\binom{n-1}{k-1}$ gives
$$
B_n(f_1)(x)=x\sum_{j=0}^{n-1}\binom{n-1}{j}x^j(1-x)^{n-1-j}=x.
$$
Thus $\boxed{B_n(f_0)=1,\quad B_n(f_1)=x}$, including $x=0,1$ by evaluation of the <polynomials>.
For $m\ge1$ and $n\ge2$, substitute the preceding coefficient identity into the <Bernstein polynomial> sum and write $j=k-1$. This yields the <Bernstein monomial recurrence>
$$
B_n(f_m)(x)=x\sum_{\ell=0}^{m-1}a_{n,m,\ell}\,B_{n-1}(f_{m-1-\ell})(x).
$$
The coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n)^{m-1}\to1$ and the others tend to zero. The weights in any <Bernstein polynomial> sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x\,x^{m-1}=x^m$: the $\ell=0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the <supremum norm> proves <uniform convergence> on $[0,1]$.
As an independent quantitative bound, take a <binomial distribution> variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb E[(K/n)^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the <Cauchy-Schwarz inequality> and the <variance> of $K/n$ give
$$
\boxed{|B_n(f_m)(x)-x^m|\le m\sqrt{\frac{x(1-x)}n}\le\frac m{2\sqrt n}.}
$$
For $m=0$ the error is exactly zero. Finally the requested sum equals $2^m B_{2n}(f_m)(1/2)$, so
$$
\boxed{\lim_{n\to\infty}\frac1{4^n}\sum_{k=0}^{2n}\left(\frac kn\right)^m\binom{2n}{k}=2^m(1/2)^m=1.}
$$
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