How much of the energy of a car is required for overcoming air resistance? How much of the energy of a car is required for overcoming air resistance? For example, assume a car going 65 mph has a mileage of 25 gpm. If all air resistance was eliminated, approximately what would be the gain in mileage?
I am trying to get a general idea of how much air resistance contributes to the forces opposing the forward motion of a car.
 A: This started as a comment but I will expand it as an answer since it warrants it. 
When you look at something like miles per gallon, you are looking at all of the total losses in the car. This is air drag, friction with the ground, and losses in the engine itself. To put it into perspective, the average efficiency of an internal combustion engine is only around 20%. So there's a lot of energy lost just in the engine.
But to answer your direct question, almost all of the opposing force on the vehicle is air drag. Friction with the ground is very small. The rolling resistance is proportional to the weight of the vehicle and for car tires the coefficient is small, we'll round and say 0.01. 
For air drag, it depends on the drag coefficient and surface area of the car. This coefficient is orders of magnitude larger than the rolling resistance. But drag also depends on the speed squared. So doubling your speed causes a force 4 times larger. This is a completely dominant effect once the speed is big enough to be interesting (a few mph).
To put some equations to it, $F_f = c_f W$ is the rolling resistance force with $c_f$ the coefficient of friction and $W$ the weight of the car. We will assume that the weight is proportional to the volume. $F_d = 1/2 \rho c_d A U^2$ is the air drag force with $\rho$ the air density, $c_d$ the drag coefficient, $A$ the surface area and $U$ the speed. 
If $W = \rho_c g V$ where $\rho_c$ is an approximate density of the car. We can approximate a car as a rectangle with height $h$, width $d$ and length $l$. This means the volume of the car is $V = hdl$ and the frontal surface area is $A = dh$ giving $A = V/l$. 
With that out of the way, let's take the ratio of the forces:
$$ \frac{F_d}{F_f} = \frac{\frac{1}{2}\rho c_d \frac{V}{l} U^2}{c_f \rho_c g V} = \frac{c_d\rho  U^2}{2l c_f g \rho_c}$$
That looks kind of confusing, so let's just look at the terms in it. If the car is roughly 2 meters wide and 5 meters long (not uncommon numbers), then we can guess that $l \approx 5$. I'm just guessing here, but I'll say $\rho_c \approx 100$ (let's say 1500 kg and using the dimensions for our cube). The density of air is around 1. So this gives $\rho/\rho_c \approx 0.01$. We know that $c_d/c_f \approx 50$ also. This gives us something that looks like:
$$ \frac{F_d}{F_f} \approx 50\times0.01\times U^2/5/10 \approx 0.1\times U^2$$
which turns out to be a really simple:
$$ \frac{F_d}{F_f} \approx 0.01 U^2 $$
which means that once you get over roughly $11~\text{m/s}$ for speed (roughly 21 mph), the air drag begins to take over as dominant. Obviously this is all approximate, but it should make it pretty clear that once a car is moving, air drag dominates.
I should note that the surface are approximation here is much larger than the real surface area. However, it doesn't change the results by all that much. The point is, at any speed you are likely to encounter outside of a parking lot, the air drag will contribute the majority of the opposition.
A: Conceptually, I think this problem is better stated in terms of work and power instead of energy. Work is the product of force and distance. Power is work per unit time. If the force from air friction is cut in half, it takes half as much power to overcome the friction. I suggest using work and power because of the way force and speed can confuse a problem. (Like if I am in a vaccuum can I pedal my bike long enough to reach escape velocity on the Moon? Why does a car stop accelerating?).
It is also pretty easy to do the experiment. In the most basic form, use a car with a manifold pressure gauge and calibrate with some hill grades and other tests. Tuck in on the rear bumper of a semi-truck trailer like a race driver drafting on the car ahead, and compare to open road conditions. On my high drag old very powerful Jimmy, the gauge dropped (rose) to nearly zero.
When enough cars are driven by non-biogical autonomous systems, I'm sure drafting will be a common way to increase efficiency. Maybe the leader will be automatically credited from breaking the way or there will be rotation as in cycling races and geese flying.
With a well instrumented car and an arrangement with a truck driver you can get very good data. Bicycle speed records have been set by drafting behind a special vehicle - and once by being enclosed by a train boxcar with a wooden roadway laid into the track.
A: Real world numbers:
Camry: 60 lbs of force pushing back at 60mph
S10:     90 lbs of force @60
Hummer:. 150 lbs @60
Good example of how much air can influence your mpg. I do know that a "bug screen" can drop 1mph or even roof racks. That is my own experience. That and after 55mpg your force becomes more restrictive. Cummins released a study on all this as well. Best of luck.
A: Back in the 1970's, when the Arabs imposed an oil embargo in the wake of the Yom Kippur war, President Nixon ordered speed limits to be reduced to 55 mph. This was to conserve gasoline. I don't remember the details, but there was some kind of optimum gas mileage at that speed.
Some losses (like engine friction and rolling friction) will tend to be linear with speed. So they don't really hurt mileage. Air friction is the main loss which is quadratic with speed. When you have two parameters, one of them linear and one quadratic, the quadratic one eventually dominates. And it's a funny consequence of the optimisation equations that the crossover point occurs when the two losses are equal. 
If President Nixon wanted us to keep below 55 mph, I'm going to understand that must have been pretty close to the speed where air friction took over as the dominant loss. So assuming that the air friction was equal to all other losses at that speed, it's easy to see that if you're getting 30 mpg at 55 mph, you'd probably be getting around 60 mpg if you were driving in a vacuum. 
