Both in practice and in theory, the best way to get a black body spectrum is to make a small hole in the wall of a hot cavity. Any light entering the hole will probably bounce around many times, and eventually be absorbed, before it could leave the cavity.
The invariance of the spectrum follows from the Second Law of Thermodynamics. (This argument doesn't give any insight to the mechanisms, but is nevertheless enlightening.) Two cavities, made of different materials but at the same temperature, face each other. All the radiation from one enters the other and vice versa. If their total emitted radiation was different, there would be a net flow of energy from one to the other - which is forbidden by the Second Law (and the Zeroth law too, for that matter). If the total radiation was the same, but the spectra were different, you could come up with a simple scheme such as placing a bandpass filter between them, to make sure that if body A emits more than body B at a specific wavelength, only that wavelength is allowed through, still causing energy to pass from A to B.
Your questions, @CuriousOne:
Is the black body radiation formula applicable to an macroscopic object composed of different elements only?
It is applicable to any body, but only if it is actually black at all wavelengths. A microscopic object can also be a black body. For example, a plasma consisting of one free proton and one electron inside a perfectly reflecting box with a small hole is a black body. (See later answer for details.)
How can it describe so different things (eg Sun spectrum, CMB) if there is nothing in the formula that relates to a specific material/element?
As long as there is some mechanism of generating photons of any required wavelength, thermodynamics will see to it that such photons are generated in the required numbers. If there is no such mechanism, the body will not be black. For example, if you found a material that was completely transparent to green light because it had no atoms capable of absorbing green light, then this body would not generate the green part of the black body spectrum - because it is not black.
The one-electron plasma I mentioned would be OK because the free electron can have any energy, and can therefore emit and absorb at any wavelength.
If the photons come from vibrational/rotational levels transitions, shouldn't these depend on the object in question?
See previous answer.
Also, what's the actual reason why there are no contributions from translational motion? And what about excited electrons?
Whenever free electrons are present, they will contribute to the emitted radiation through their translational motion. In a plasma, this mechanism is called bremsstrahlung. In a metal, it is called reflection. In an atom, emission can only occur if an electron is excited. An atom in the ground state cannot emit. A molecule can emit even if there is no electronic excitation, because it also has vibrational and rotational levels.