NDSolve with vectors

Mathematica doesn't have vector variables (yet). That is to say, you can assign a list to a variable, but you cannot use a variable in a function like NDSolve and let Mathematica work out its dimensions or let the dimensions be undetermined.

If you change your function to this:

gravity = 10;
withDrag[p0_, v0_, drag_] :=
 Module[{p},
  p[t_] := {p1[t], p2[t], p3[t]};
  p[t] /.
   NDSolve[
    Thread /@ {
       p[0] == p0,
       p'[0] == v0,
       p''[t] == drag*Norm[p'[t]]*p'[t] + {0, 0, -gravity}} // Flatten,
    p[t],
    {t, 0, 5}
    ]// First
  ]

it works. What is does is defining your p as a vector (list) of functions. Thread takes care of distributing == over the vector components and Flatten makes a single list of equations from all this.

track[t_] = withDrag[{0, 0, 0}, {0, 10^2, 10}, 0.001];

ParametricPlot3D[track[t], {t, 0, 5}, BoxRatios -> 1]

Note that I reduced the starting value of v0[[2]] to 10^2 because 10^4 yields a 'stiff' system. Also note that I used BoxRatios -> 1 to prevent the box from becoming flat.

While under the hood this method still provides Mathematica with the 9 equations that you already tried manually, it has the advantage that it leaves your vector equations intact.

As of Version 9, you can work with vectors in NDSolve[]!:

gravity = 10;
withDrag[p0_, v0_, drag_] := Module[{p},
  p[t_] := Evaluate@Array[Unique[][t] &, 3];
  p[t] /. NDSolve[{
                   p[0]   == p0,
                   p'[0]  == v0,
                   p''[t] == drag*Norm[p'[t]]*p'[t] + {0, 0, -gravity}},
               p[t], {t, 0, 5}] // First]

track[t_] = withDrag[{0, 0, 0}, {0, 10^2, 10}, 0.001];
ParametricPlot3D[track[t], {t, 0, 5}, BoxRatios -> 1]

Having a helper function rhs, which evaluates only with a numeric vector as argument, for the right-hand side of the force equation lets you use vectors as you want. This way the undesired symbolic precalculation (threading of drag (v.v) Normalize[v] with {0, 0, gravity}) is bypassed and the solving continues numerically. See this answer for a bit more detail.

Physically, the drag term should be negative. Also, just as an interesting angle, I added WhenEvent "equation" that terminates the integration when particle hits the ground.

withDrag[p0_, v0_, drag_] :=
 Module[{gravity = 10, rhs},
  rhs[v_?(VectorQ[#, NumericQ] &)] :=
   -drag (v.v) Normalize[v] - {0, 0, gravity};
  NDSolveValue[{
    p''[t] == rhs[p'[t]],
    p'[0] == v0,
    p[0] == p0,
    WhenEvent[p[t][[3]] < 0, "StopIntegration"]}, 
   p, {t, 0, \[Infinity]}]]

The solution time depends on the initial values, it can be extracted with suitable parting.

sol = withDrag[{0, 0, 0}, {10, 10, 100}, .1];

ParametricPlot3D[sol[t], {t, 0, sol[[1, 1, 2]]},
 BoxRatios -> 1,
 ImageSize -> Small]