First, I want to point out some of the things I will be using and how I will be notating.
Bold indicates that the unit given is a Euclidean vector.
i, j, k represent unit vectors in the x, y and z axis, respectively.
^ indicates a vector cross product (Powers are written in ^superscript).
_Subscript denotes the part of a vector, IE, F_x would be the force in the x direction.
|symbol| indicates a magnitude.

Lets assume the trike can be represented as a simple triangular prism with the wheels at each edge. It has a mass m and is travelling at velocity v in a circle of radius R. For mathematical simplicity the centre of mass is located halfway along the centre line and halfway up.

The axis of rotation in this case is given by the red line. The coordinate axis have been defined as along the axis (z), parallel to the floor and perpendicular to the axis (x) and perpendicular to both (y). As a note, I have defined the CoM as l from the back along the centre line and h from the floor, the actual vehicle can be any size, such as 72l and 65h instead of 2l and 2h, all we are concerned with is the CoM, it's just easier to visualise if it's about centred.

Friction is assumed to be "infinite" in the x direction so that the vehicle does not slide (infact, it doesn't need to be infinite, just equal to or greater than the centripetal force).

First, we get the magnitude of the centripetal force, using the formula for circular motion;

|F|=mv²/R

This force is pretty useless to us, as we don't know how it behaves in our coordinate axis system. This force acts perpendicular to our centre line, which is at Θ^o to our z axis, and parallel to the floor. This gives;

F=(mv²/R)cosΘi+(mv²/R)sinΘk

As we are dealing with stuff rotating around z, we can ignore the k component that acts along z. This allows us to simplify to;

|F_x|=(mv²/R)cosΘ

While this is generally considered the inward force required to create such a circular motion, you can also view it as a force exerted outwards on the system. Of course, this force acts upon the centre of mass.

Using this force with the equation for torque (τ=r^F, where r is the radius from the centre of motion) we can find the rotational force (torque) exerted on the trike by the turning.

By virtue of the cross product, it turns out that it doesn't matter how far away the CoM is laterally, only vertically. This gives;

|τ_CW|=(mv²/R)hcosΘ

Now we need to know the torque due to gravity, which is calculated in a similar way. First we work out the force due to gravity;

F=mg

The direction of g (and therefore F) is directly down, IE -j.

The distance r in this case is found using simple trig to be lsinΘi.

This gives us a torque;

|τ_CCW|=mglsinΘ

Now, whether the vehicle will roll or not is simply a matter of which torque is greater, it will roll when;

|τ_CCW|<|τ_CW|

Which can also be written as;

mglsinΘ<(mv²/R)hcosΘ

Interestingly enough, the masses cancel in this. While this may seem wrong due to common sense, it isn't. In actual fact, changing the mass changes the moment of inertia of the vehicle. This means that while a heavier vehicle will just as likely roll, it will take a lot longer to do so as it accelerates slower. If we move everything to one side, we can see that it will roll if;

(Rgl/hv²)tanΘ<1

This means that decreasing R, g, l and Θ makes it more likely to roll, while increasing h and v does the same. Bear in mind that Θ=0 is the equivalent of a bike, not a car. A car would need a similar approach to the above, but would yeild a result based on its width and not the front/rear placement of the CoM.

It's also worth noting that having the vehicle the other way round will yeild exactly the same result, so neither way round is more stable than the other, until you start taking into account where acceleration of the vehicle is applied, but those are inconsequential for a basic model of cornering. Also, suddenly turning into a corner would cause some other forces to start acting, which would also change thing, but in a smooth turn in, these would be negligible.

Hope you enjoyed that and I apologise for any errors, I cba to check it over