If you take any of the platonic solids. Take a random corner, don't worry they're all the same, and make this the center of a sphere, where the radius is the distance from your chosen corner to the farthest corner from it. So on a cube the radius would be $\sqrt3\phantom.\cdot\phantom.${sidelength}. Do this for all the corners. The part where all the spheres overlap is a solid of constant width.
Am I correct?
It turns out, you are not!
More specifically, what you're talking about is a generalization of the so-called Reuleaux triangle, and in the simplest case — namely, the one where the solid is a tetrahedron — the resulting surface isn't constant-width; the problem is that an edge-to-edge span can be greater than the distance to a vertex. For more details, see https://en.wikipedia.org/wiki/Reuleaux_tetrahedron .