Let's consider an icosahedron with edges of length L = 1. Later we can multiply by L^2 or L^3 for the area and volume respectively.

Each of the 20 faces is a regular triangle, with base 1 and height sqrt(3)/2, for an area of sqrt(3)/4.  There are 20 of them so the total area of this unit icosahedron is 5*sqrt(3).  For the general icosahedron it's 5 * L^2 * sqrt(3).

To get the volume we'd need the perpendicular distance from the center of the icosahedron to one of the faces, for use of that face as the base of a pyramid. This would be half the distance between the planes of the opposite-face triangles, even if the line from the center is not itself perpendicular to the triangular base; the between-plane distance is twice the height of the pyramid.

When looking at the icosahedron with one's eye aligned with a vertex, the center and the opposite vertex one sees a regular pentagon surrounding the near vertex. The sides of the pentagon are of course unit length. The outer triangle (i.e., not one of the triangles internal to this pentagon) sharing an edge with this pentagon is parallel to its opposite face on the icosahedron, and that opposite face shares a vertex with the pentagon, opposite the edge previously mentioned.

The distance from the vertex to that opposite edge is 1/(2*tan(36°)) + 1/(2*sin(36°))  ~=  1.538841768587627. However this is not the distance between the two opposite faces, as the faces are slanted relative to the perpendicular to the plane of the pentagon we've been talking about. And there doesn't seem to be an obvious relationship between this slant angle and the dihedral angle of the icosahedron.