However, this comes to odds with the idea most people have of metals, which is a collection of shiny (highly reflective) grey solids. This is because while they can absorb photons of almost arbitrary wavelength, the absorption probability is not as high as in other materials. As such, a secondary effect can take place where the incoming photons interact with the free electrons in the metal and cause the light to be reflected back, rather than absorbed. One simple way to decrease reflection and hence turn pure metals into black substances is to grind them into a very fine dust or otherwise make a very rough surface on the microscopic level, which causes incident light to bounce around more times when hitting the material, increasing the odds of the light being absorbed.

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Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound.

The record is currently held by Vantablack[1], a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black.

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Furthermore, while we often think of metals as materials which are composed chiefly of atoms from the $s$/$d$/$f$/early $p$ blocks of elements from the periodic table (the "metallic elements"), metals are actually not defined by their composition at all, but by their electronic structure. It is even possible to have metals made out of "non-metals" such as pure carbon, for example (carbon nanotubes of certain chiralities). Thus metallic character is actually likely a contributing factor to the darkness of Vantablack, which bon mentions in his answer.

All metals are capable of absorbing photons of any wavelength below hard ultraviolet, as ideally there are allowed electronic transitions of arbitrarily small energy between states in the unfilled valence band. This means metals are theoretically the materials with the widest wavelength range for photon absorption (except for plasmas).

Any black compound absorbs photons in all the visible spectrum; that is why essentially it appears black to our eyes. So, for instance, iron(II,III) oxide, $\ce{Fe3O4}$ will do so; it is even used as a black pigment. It is very unlikely though that it will absorb each and every photon of any wavelength in the visible spectrum, but the search for such a perfect absorber is meaningless in any case since the very notion of a visible spectrum is not strict.