[John]

I assumed that the CoG is halfway between the front and back, if that is false pretty much none of this makes sense.

Fn(max) = Nn*mu

Since its symmetrical:
N1 = N2
N3 = N4

summing forces
W = N1+N2+N3+N4 = 2*N1 + 2*N3
summing moments about axis of front axles, defining wheelbase as 2 units
W*(1+f1) = (N3+N4)*2 = 4 * N3
combining formulas
W= 2*N1 + W*(1+f1)/2
W(1-1/2-(f1)/2) = 2*N1
N1 = W(1-f1)/4

F1(max) = F2(max) = mu*W(1-f1)/4
F3(max) = F4(max) = mu*W(1+f2)/4

summing moments around vertical axis through CoG
0=F1y*1+F2y*1+F3y*1+F4y*1-F1x*f2-F2x*f2-F3x*f2-F4x*f2
( F1x + F3x )*f2 = F1y + F3y

by Pythagorean theorem:
(sqrt((F1)^2-(F1y)^2) + sqrt((F3)^2-(F3y)^2))*f2 = F1y + F3y

Fny = tau/r

(sqrt((mu*W(1-f1)/4)^2-(tau/r)^2) + sqrt((mu*W(1+f1)/4)^2-(tau/r)^2)) *f2 = 2*tau/r

I spent around half an hour trying to simplify that but just made it worse:

sqrt((mu*W(1-f1)/4)^2-(tau/r)^2) * f2 = 2*tau/r - sqrt((mu*W(1+f1)/4)^2-(tau/r)^2)) *f2

(mu*W(1-f1)/4)^2-(tau/r)^2) * (f2)^2 = 4 * (tau/r)^2 - 4 * tau/r * sqrt((mu*W(1+f1)/4)^2-(tau/r)^2)) *f2 + ((mu*W(1+f1)/4)^2-(tau/r)^2)*(f2)^2

4*tau/r * sqrt((mu*W(1+f1)/4)^2-(tau/r)^2)) *f2 = 4*(tau/r)^2 + (f2)^2 * (((mu*W)^2*f1)/4)

16 * (tau/r)^2 * ((mu*W(1+f1)/4)^2-(tau/r)^2) *(f2)^2 = 16 * (tau/r)^4 + 8*(tau/r)^2 * (f2)^2 * (((mu*W)^2*f1)/4) + (f2)^4 * (mu*W)^4 * ((f1)^2)/16