Cube using single GL_TRIANGLE_STRIP

For those of you who are lazy (like me), here's a copy-paste version of rob mayoff's answer ;)

static const GLfloat cube_strip[] = {
    -1.f, 1.f, 1.f,     // Front-top-left
    1.f, 1.f, 1.f,      // Front-top-right
    -1.f, -1.f, 1.f,    // Front-bottom-left
    1.f, -1.f, 1.f,     // Front-bottom-right
    1.f, -1.f, -1.f,    // Back-bottom-right
    1.f, 1.f, 1.f,      // Front-top-right
    1.f, 1.f, -1.f,     // Back-top-right
    -1.f, 1.f, 1.f,     // Front-top-left
    -1.f, 1.f, -1.f,    // Back-top-left
    -1.f, -1.f, 1.f,    // Front-bottom-left
    -1.f, -1.f, -1.f,   // Back-bottom-left
    1.f, -1.f, -1.f,    // Back-bottom-right
    -1.f, 1.f, -1.f,    // Back-top-left
    1.f, 1.f, -1.f      // Back-top-right
};

May be useful to some, here's a geometry shader that will take in a point and output a triangle strip of a unit cube

#version 410

layout(points) in;
layout(triangle_strip, max_vertices = 12) out;

uniform mat4 mvp;

void main() {
    vec4 center = gl_in[0].gl_Position;

    vec4 dx = mvp[0];
    vec4 dy = mvp[1];
    vec4 dz = mvp[2];

    vec4 p1 = center;
    vec4 p2 = center + dx;
    vec4 p3 = center + dy;
    vec4 p4 = p2 + dy;
    vec4 p5 = p1 + dz;
    vec4 p6 = p2 + dz;
    vec4 p7 = p3 + dz;
    vec4 p8 = p4 + dz;

    gl_Position = p7;
    EmitVertex();

    gl_Position = p8;
    EmitVertex();

    gl_Position = p5;
    EmitVertex();

    gl_Position = p6;
    EmitVertex();

    gl_Position = p2;
    EmitVertex();

    gl_Position = p8;
    EmitVertex();

    gl_Position = p4;
    EmitVertex();

    gl_Position = p7;
    EmitVertex();

    gl_Position = p3;
    EmitVertex();

    gl_Position = p5;
    EmitVertex();

    gl_Position = p1;
    EmitVertex();

    gl_Position = p2;
    EmitVertex();

    gl_Position = p3;
    EmitVertex();

    gl_Position = p4;
    EmitVertex();

}

Yes, after a bit of experimenting I found the answer myself. Imagine the corners of your cube are colored alternatingly black and white. Draw a triangle edge along each face between the two black corners. That way, the diagonals form a tetrahedron inside the cube. For the [0,1]³ cube, a possible sequence of coordinates would be the following:

Vertex  Triangle    Face
------+-----------+-----
0 0 0
0 1 0
1 0 0  000 010 100  **0
1 1 0  100 010 110  **0
1 1 1  100 110 111  1**
0 1 0  111 110 010  *1*
0 1 1  111 010 011  *1*
0 0 1  011 010 001  0**
1 1 1  011 001 111  **1
1 0 1  111 001 101  **1
1 0 0  111 101 100  1**
0 0 1  100 101 001  *0*
0 0 0  100 001 000  *0*
0 1 0  000 001 010  0**

From the paper Optimizing Triangle Strips for Fast Rendering by Evans, Skiena, and Varshney: