Why does the wavefunction of a particle spread out after a measurement?

It is a mathematical phenomenon, it happens due to time-dependent Schroedinger equation. It is similar to diffusion equation, localized peaks in tend to dissipate according to both equations. So any localized wavepacket spreads unless that spreading is stopped by a potential rise.

This behaviour can be most easily shown and quantified for free particle. If particle is in some potential well, that well will slow down or even stop the spreading.

Assuming by "spread" you mean how "wide" the distribution of $mid psi(x,t) mid^2$ (or for momentum space, the spread of $mid psi(k,t) mid^2$).

Quantum mechanics states that the wave packet of a particle "spreads-out" in position again after a measurement on this particle has been made.

Before and after measurement, the system evolves as described by the non-relativistic time-dependant Schrodinger equation $$ifrac{ partial psi(x,t)}{partial t}=hat{H} psi(x,t)$$

The initial conditions for the system after the first measurement, are dependant on how you performed this measurement that caused the wave-function to collapse. So if you make the exact same measurement the same way, a very short time after the first, you will get a similar result to the first measurement. If you measured position, you will get sharply peaked distribution (Dirac delta function) in position space $delta (x - x')$ centred at $x'$ causing a "spread", or a large uncertainty in the momentum distribution. The wave packet does become wider.

Is this spreading or "dispersion" caused by inherent properties of the wavefunction itself (eg according to the Uncertainty Principle), or is it due to the interaction with the environment (or both)?

The wavefunction on its own has no physical properties. It is its square that has physical meaning. The uncertainty principle tells us that the more accurately we measure position, the less accurate we can know its momentum, and vice-versa. How it is measured (with what apparatus, how we do it etc) can affect the accuracy we get for either quantity, but this complementarity between accuracy of position/momentum is still described by the uncertainty relation. So the environment or apparatus is important only in terms of accuracy, which will obviously affect the amount of "spread".

Does the wavefunction of a free, non-interacting particle also show this behaviour?

Yes it does. The Heisenberg uncertainty relation still holds. And solving the Schrodinger equation will give you Gaussian distributions in both position and momentum space. Again, in free space, if you describe the particle with a wave packet, and you wanted to increase the accuracy in momentum, the position uncertainty does become wider.

Quantum mechanics states that the wave packet of a particle "spreads-out" in position again after a measurement on this particle has been made

Quantum mechanics is a mathematical theory which developed over many decades in order to mathematically model the observations and data mainly in the dimensions of microns and nanometers, so as to be able to predict experiments. It depends on postulates , part of which are the existence of quantum mechanical wave equations, but there are different mathematical models based on these postulates according to the complexity and need of the data at hand.

The ultimate calculation tool based on quantum mechanics is quantum field theory, QFT, which can model data from particle physics interactions to superconductivity. So your statement "Quantum mechanics states" is not a general postulate statement of quantum mechanics but of a particular model, a primitive model as far as particle physics goes.

Wavefunctions have the properties needed to model a specific situation.

Particle physics uses QFT to model the data, which means interactions and decays can be calculated using Feynman diagrams. The modeling of the particle in a Feynman diagram does not have a spread in the wave function, as the particles are described by creation and annihilation operators on plane-wave wavefunctions of specific energy and momentum. The wavepacket description of a free particle in vacuum comes if one wants to model it in QFT states. Plane waves give a probability for the particle to be anyplace in space-time in QFT creation and annihilation operators, the wavepacket localizes in space-time the probability of finding the particle.

Does the wavefunction of a free, non-interacting particle also show this behaviour?

The wavepacket solution of wave equations brings consistency to the model with QFT. It is not particularly useful because calculations for particle interactions are done in the energy-momentum variables, where a specific momentum and energy vector is assumed, not a wavepacket one. This mathematical modeling is successful the predictive power of the standard model ( a QFT theory), is continually validated.

In conclusion, one should use the particular quantum mechanical mathematical model for the particular experiment.

The space wavepacket model representing a free particle spreads anyway, without interacting see here .