The case of the cube corner is obvious: 1/8.

The puzzle is really about the octahedron.

At each corner of the octahedron, four equilateral triangles meet. The triangular faces intersect the surface of the sphere in great circle arcs of 60° each. Thus four of these arcs form a spherical square with each side being a great circle arc of 60°. The angles of the square are larger than 90°, and are the dihedral angles of the octahedron. The excess of the total measures of these angles over 360° is a measure of the area.

To get an idea of how the angular excess measures the area, consider the case of the cube in the first part of this puzzle. The dihedrals of the cube are all 90°. Considered as the angles of the intercepted spherical triangle the three add up to 270°, or an excess of 90° above the total for a plane triangle, and that gave us 1/8 of the spherical surface. We'll use that ratio later.

Back to the octahedral case: Consider four triangles made by constructing two diagonals of the spherical square. Now consider one of these isosceles triangles and divide it down the middle to make a right triangle on each side.

One leg of this triangle is 30°, as it's half one of the 60° sides of the square. The bisector we drew meets the base at 90°, so it's a right triangle with one leg of 30°. Another angle, opposite the 30° side, is 360/8 = 45°, being half of one of the quadrant angles meeting at the center of the square. Now we have to solve this triangle from angle-side-angle without being the case that the angles of a triangle add to 180° -- they don't in spherical trig, as the excess is just what we're using to get the area.

The needed version of the spherical law of cosines is

cos(A) = - cos(B) cos(C) + sin(B) sin(C) cos(a)

cos(45°) = - cos(90°)*cos(A) + sin(90°)*sin(A)*cos(30°)

cos(45°) = sin(A)*cos(30°)

 

  sin(A) = cos(45°)/cos(30°)

  

A is half the dihedral angle of the octahedron, but we need four times the dihedral angle, or 8*A to see how much excess it is above 360°.  

A is about 54.7356103172454°. 8*A - 360 =~ 77.8848825379628°.

Remember that an excess of 90° gave an area of 1/8 of the spherical surface. So:

(1/8)/90 = x/77.8848825379628

x = (77.8848825379628/8)/90 =  0.108173447969393 = 1/9.24441273502668 of the complete spherical surface.