Why is the magnitude of friction acting on the rear wheels greater than the magnitude of friction acting on the front wheels?

I think you have misunderstood how friction slows cars down. The friction between the tires and road allow the car to move (think about it). The friction between the axles and wheel are what slow the car down.

Let's look a the case of rear wheel drive and designate all the forces acting during acceleration (air drag, ball bearings friction and other non-conservative forces are ignored)

In the $y$-direction where there's no acceleration we get:

$$N_F+N_B-mg=0tag{1}$$

In the $x$-direction:

$$ma=F_B-F_F$$ This the Equation of Motion of the car.

---

Readers may wonder why $F_F$ points in the $-x$ direction. When the car accelerates, friction is needed on the front wheels to sustain the increase in angular velocity $omega$.

For rolling without slipping in uniform motion (constant velocity), with $R$ the wheel's radius:

$$v=omega R$$

In acceleration, rolling without slipping:

$$a=frac{mathbf{d}omega}{mathbf{d}t}R=dot{omega}R$$

To get this angular acceleration clockwise torque $tau$ needs to act on the wheel:

$$tau=Idot{omega}$$

This torque is provided by the friction force $F_F$, so that:

$$tau=F_F times R$$

---

The balance of torques about the CoG must be zero, to prevent the car from starting to rotate, so:

$$N_BL_1+F_F h=N_F L_2+F_Bh$$

And with: $F_F=mu N_F$ and $F_B=mu N_B$, then:

$$N_BL_1+mu N_F h=N_F L_2+mu N_Bh$$

$$N_B(L_1-mu h)=N_F(L_2-mu h)$$

Combined with $(1)$ and some reworking we get:

$$N_B=frac{L_2-mu h}{L_1+L_2-2 mu h}tag{2}mg$$ $$N_F=frac{L_1-mu h}{L_1+L_2-2 mu h}tag{3}mg$$

Now compare this to the static case, where $a=0$ and $v=text{constant}$. It can beshown easily that in that case:

$$N_B=frac{L_2}{L_1+L_2}mgtag{A}$$ $$N_F=frac{L_1}{L_1+L_2}mgtag{B}$$

Comparing $(2)$ to $(B)$ it is clear that in the case of the accelerating car, the normal force on the rear wheels and thus also the friction, is higher than in the static ($a=0$) case.