How can I calculate the force on the nose gear at landing?
Calculating the loads on the nose and main landing gear isn't really different. The answer you linked is a highly simplified approach you should be able to use for the nose landing gear too, but I will provide a "less simplified" approach to calculate the vertical forces. The braking force is already greatly described there.

A simplified approach

First you need to know the impact speed and the spring and damping characteristic of your landing gear.

At touch down there are three possible cases:


Main landing gear touches the ground first. Then the airplane rotates down with an angular speed $\omega_L$. The impact speed of you nose landing gear is therefore $$V_{Impact} = \omega_L \cdot d $$ $d$: distance between the 
main and the nose landing gear.
Nose landing gear touches the ground first. (For now empty, if I get time I'll write about it later)
Both landing gears touch the ground at the same time. For a glide angle $\gamma$ the impact speed will be: $$V_{Impact} = V\cdot \sin(\gamma)$$ $V$: airplane speed


The stiffness ($k$) and damping characteristic ($c$) of the landing gear depend mainly on the suspension and the tires used. Tires inflated with a wrong pressure can be really dangerous as they can increase the load on the landing gear.

Putting this data together into a mass spring damper model like in the image



gives you an equation for the nose landing gear movement: $$m\cdot \ddot{x}+c\cdot \dot{x}+k\cdot x = -m\cdot g$$
For the reduced mass $m$ in the model use the mass carried by the nose landing gear.
$$m = LM \cdot \frac{d-d_{CG}}{d}$$
$d_{CG}$: distance from the nose landing gear to the center of gravity

$LM$: aircraft landing mass

$g$: gravity constant

The initial condition for the differential equation is: $$\dot{x}(t=0) = -V_{Impact}$$
The impact occurs at $t=0$.

To calculate the force on the nose landing gear you just have tofind the maximum acceleration and add the static and dynamic loads: $$F_{max}=-m\cdot g -m\cdot \ddot{x}_{max}$$

Real world problems

While the calculations above works for really stiff wings, in reality the loads on the landing gear will be (probably) higher. This is due to the inertial forces of the oscillating aircraft structure.

If you don't understand where this forces come from try moving your arms up and down while standing: you will feel small forces on your feet. Now repeat the experiment with two 60t heavy wings as in the A380 and you'll see that the forces are not neglectable. And were do the oscillations come from? The same as with a guitar string: after a small excitation (plucking the strings/ landing impact) the structure (guitar string/ aircraft wings) begins to oscillate.

What other things doesn't this model take into account? Coupling between nose and main landing gear, rotation inertia, nonlinear tire characteristics,...