Yes of course. And even with nose wheel brakes the main braking force would still be generated by the main wheels, so a substantial pitching moment is generated.
Now for a crude first-order approximation. We have an answer about the braking forces here, and using those 235 kN per wheel (the 747 has 16 main wheels) means that the center of gravity, which I guess to be at least 6 m above the ground, experiences inertial loads of 3,76 MN. This needs to be reacted in a moment around the main wheels with the lever arm of the wheel base. This is conveniently displayed in the drawing below and is 25.6 m. Thus, the reaction moment around the main wheels requires a nose wheel load of 881 kN in addition to the regular load of the nose wheel (which is maybe 5% of the aircraft's gross weight or 196 kN).
Boeing 747 side view (picture source)
Since the highest friction coefficient is reached at low speed, wing lift or tail downforce will not play a significant role. Braking to a full stop after an aborted take-off at full gross weight will load the nose wheels with a multiple of the static load of more than four.
In a normal landing, the forces involved are much lower and the multiple is close to one. This is really a quick back-of-the-envelope calculation, but the magnitude should still be right.