= Solution
Every <derivative> of order $n+1$ of $\sin2x$ has magnitude at most $2^{n+1}$. Pair the <Symmetric node pairing bound for interpolation on the unit interval>[symmetric nodes] $a=i/n$ and $1-a$. For $0\le x,a\le1$,
$$
|(x-a)(x-(1-a))|=|(x-1/2)^2-(a-1/2)^2|\le1/4.
$$
If there is an unpaired middle node, its factor has magnitude at most $1/2$. Thus the full nodal product is at most $2^{-(n+1)}$, giving the stronger bound $\|f-p_n\|_\infty\le1/(n+1)!$. In particular this implies the requested
$$
\boxed{\|f-p_n\|_\infty\le\frac{2^{\lfloor(n+1)/2\rfloor}}{(n+1)!}.}
$$
The bound tends to zero by factorial growth, so $\boxed{p_n\to f\text{ uniformly on }[0,1]}$. This conclusion uses smoothness of this particular function and does not assert <uniform convergence> of equispaced interpolation for arbitrary <continuous> data.
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