As Armitage stated, you're really not losing any energy due to friction with the surface of the road... there may be a very slight amount of slippage, but I highly doubt it's even close to significant.
HOWEVER, do lose energy to "friction", caused by the constant deformation of the tires so that the "patch" of the tire is flat against the surface of the road. The lower the pressure of the tires, the larger the patch is going to be. Also, them more the car weighs, the larger the patch is going to be. Thus, if the tire pressure remains constant, adding mass to the car is going to increase the rolling resistance, and thus decrease fuel economy.
Perhaps what your friend is thinking about is that with a given shape of vehicle, the force of air resistance is going to be the same, regardless of the mass of the car. However, using F=ma, the higher the mass, the lower the acceleration. Thus, a heavier vehicle won't lose speed as quickly due to air resistance. To understand this better, drop two identically sized balls - one lead, and one styrofoam. Notice that the styrofoam ball takes longer to get to the ground. However, if both were dropped in the absence of air, they'd both hit the ground at the exact same time.
So, this seems like heavier would be better.
However, because KE = 1/2 m v^2, and conservation of energy,
having a heavier car is going to require more energy to get up to a certain speed.
Since the air resistance is a force acting in the opposite direction of the motion of the car, the engine needs to provide a force (via the road/tire interface) equal in magnitude to the wind resistance to maintain a constant speed (net force = 0)
Both act on the car over the same distance, so using work = force * distance,
the amount of work the engine does has to be equal to the amount of work done to the car by the air resistance (= force * distance), else, the car will slow down. Therefore, regardless of the mass of the car, if the shape is identical, the work done against air resistance is the same.
So, the heavier car will use more energy because of increased tire deformation.
This also points to why it is so important to watch your tire inflation pressures on a regular basis... properly inflated tires can make a tremendous impact on miles per gallon; far more of a difference than typical weight differences may cause.
Want to check this hypothesis? It's pretty simple to do. Road trip! Travel 200 miles to a distant Home Depot or Lowes or some such place. Carefully measure mpg on the trip. Purchase 10-15 80 pound bags of concrete. Drive back home, avoid pot holes! Carefully measure mpg on the return trip. Even if you have a head wind on the way out, and a tail wind on the way back, I don't think it'll make that much of a difference compared to the difference caused by the weight difference. But, then again, you can even try to account for this in your experiment by returning the concrete to HD, then driving back home again.