How to measure the Force (in Newtons) produced by a solenoid?

Use the expression for the Lorentz force. If the object is a punctual object (i.e. a charged particle) the Lorentz force is

$$vec{F}=q(vec{E}+vec{v}timesvec{B}).$$

If the object is extended then you'll have to integrate

$$vec{F}=int(rhovec{E}+vec{j}timesvec{B})d^3x.$$

Notice that the object needs to be charged in order to experience an electromagnetic force. The magnetic field inside a very long solenoid is indeed $B=(μNI)/L$ and it goes parallel to the axis of the solenoid. So I think the only thing left is to ask yourself: is there any $vec{E}$ in this scenario?

A current carrying solenoid will exert a force on a ferromagnetic material which is just outside the end of the solenoid. A calculation would be complex. The magnitude and direction of the force on each small segment of the material will vary from point to point and depend on the divergence of the field and the existing or induced magnetization within the material. (Your formula is accurate only inside a long narrow solenoid.) A uniform magnetic field will not exert a net force on a magnetic dipole (bar magnet or atomic). (Solenoids are often used to shift the position of a ferromagnetic bar to activate a valve or switch.)

OK, you mean calculate. (You would measure it with a spring balance - easy).

You want to start with $H$ rather than $B$, but that's just a factor of $mu_0$.

Then you need the magnetisation $M$. For dia/paramagnets this is just $chi H$, where $chi$ is the magnetic susceptibility. For ferromagnets it's more complicated due to saturation and hysteresis.

If the object is large and/or susceptible (ferromagnets again) you have to include the effect of the magnetised object on the external field and on itself and calculations get complicated. So let us suppose it is small enough not to disturb the picture.

Then the energy is just $-M.B$ integrated over the volume, whatever the shape is. $M$ and $B$ are colinear here so you don't need to worry about the dot product.

Having got that, the force is just the gradient of the energy and can be calculated as a function of position if you know the field. There is no force in the barrel of the solenoid where the field is uniform, but it appears at the ends where the field lines diverge.