Answer: no, there does not exist such a polyhedron. First let us prove the following Lemma. The area of a face $\mathcal{F}$ of polyhedron $\mathcal{P}$ is less than the sum of the areas of the other faces. (In fact the strictness of the relation is not needed for solving the problem.) Proof. Draw a plane throw each side of $\mathcal{F}$ perpendicular to its plane. Name by $\mathcal{S}$ the slice formed by these planes. Every face of the polyhedron $\mathcal{P}$ or a part of it lies in $\mathcal{S}$. Thus because area of a figure is not less than its projection (on the plane of $\mathcal{F}$) then the area of the part of the surface of $\mathcal{P}$ lying inside $\mathcal{S}$ is not less than the area of $\mathcal{F}$. If there is a part of $\mathcal{F}$ lying outside of $\mathcal{S}$ then the lemma is proven. Otherwise there should be a face which is not parallel to $\mathcal{F}$ thus the area of its part lying in $\mathcal{S}$ is strictly greater than the area of its projection on $\mathcal{F}$ and again we finish the proof of the lemma. Now suppose there exists a polyhedron satisfying the condition of the problem. Let $S_1<S_2<…<S_n$ be the areas of all the faces of the polyhedron. Then $S_1+S_2+…+S_{n-1}>S_n\ge 2S_{n-1}\ge S_{n-1}+2S_{n-2}\ge … \ge S_{n-1}+S_{n-2}+…+S_2+2S_1$ — contradiction.