In a sense made precise below, the limit can be within $\varepsilon$ of any convex polytope.



- Suppose one takes at each stage not just the midpoints of the edges but $\frac{P+Q}2$ for *every* pair of points. Then one will get the same convex hull at each stage, there will just be points other than the vertices included.

- Start with a polytope $P$ then replace each vertex $v$ by three vertices $v_1,v_2,v_3$ forming a   triangle with sides of length (less  than) $\varepsilon$ with centroid $v$ and containing no point of $P$ other than $v.$ This gives a polytope $P_1 \supset P$ with three times as many vertices but with all points within $\varepsilon$ of $P.$ Then as the process is applied to $P_1$ one will get a nested series of polytopes all containing $P.$ However the limit is larger than $P$: If $v,w$ are vertices of $P$ then the limit will contain $\frac{v_1}{2}+\frac{w}{2}.$ (At stage $k$ we will have $\frac{v_1}{2}+\frac{a}{2^k}w_1+\frac{b}{2^k}w_2+\frac{c}{2^k}w_3$ with $a+b+c=2^{k-1}$ and nearly equal)

This still leaves me a bit short of a description of the exact limit for a specific polytope. If $P_1$ has $n$ vertices $V_1,V_2,\cdots$ then it should be all points $W=\sum c_i V_i$ subject to some inequalities. Certainly the points $c_iV_i+c_jV_j+c_kV_k$ with $\max{c_i,c_j,c_k} \le \frac13$ are in the limit. If  a point $W$ has $c_1=0$ then $\alpha V_1+(1-\alpha) W$ is in the limit for any $\alpha \le \frac 12.$