All the answers name some valid points, but they fail to really answer the question which I want to repeat for clarity:
Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?
The Short & Simple Answer
These two impedances have no relation at all. They describe different physical phenomena: the antenna input impedance is not related to the 377 Ohm free-space impedance. It is only "by accident" that the unit of both terms is the same (i,e., Ohms). Furthermore, 50 Ohm is just a common value for line impedances etc., see the other answers.
Basicallym the input resistance (and transmission lines impedances) is for voltages and currents, while the free space wave impedance is for electric and magnetic fields.
The Longer Answer
The first impedance mentioned in the question is the input impedance of the antenna, which is a sum of radiation resistance and losses. It is related to currents (I) and voltages (V) on a circuit-description level, i.e., $$R = \frac{V}{I}\,.$$ This impedance of the resistor is the same kind as the the transmission line impedance of coaxial lines or microstrip lines, since these are also defined via voltages and currents.
The second impedance is a wave impedance of the fields, which describes the ratios of electric (E) and magnetic (H) fields. The free space impedance, for instance is given as $$ Z_{0,\mathrm{free\,space}} = \frac{E}{H} = 119,9169832\,\Omega\,.$$ We can immediately see that fields and voltages have a relation that might change with geometry etc, or there might be no unique definition of voltages (e.g., in a hollow waveguide).
To make this lack of relation of these kinds of impedances more clear, an example might help. In the very simple case of the TEM wave inside of a coaxial cable, we know how to calculate the transmission line impedance based on the geometry as $$Z_{0,\mathrm{coax}}=\frac{1}{2\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\ln\frac{D}{d}\,,$$ if we assume that the filling material is vacuum. This is a transmission line impedance for the currents and voltages of this line, and this is the line impedance which should be matched to the input impedance of an antenna.
However, having a look at the fields inside the cable, we find that the electric field has only the radial component (exact values are irrelevant in this context) $$E_r \propto \frac{1}{r \ln(r_{\mathrm{inner}}/r_{\mathrm{outer}})} \,.$$ More interestingly, the B field has only a phi-component which is a scaled version of the electric radial field $$B_\phi = \frac{k}{\omega}E_r=\frac{1}{c}E_r\,,$$ where c is the speed of light, which is from free space (!) because the medium inside is free space. By using $$ B = \mu H\,,$$ we finally know the phi-component of the magnetic field as $$H_\phi =\frac{\sqrt{\epsilon}}{\sqrt{\mu}}E_r=Z_{0,\mathrm{free\,space}}E_r\,,$$ Therefore, the ratio of electric and magnetic fields is constant and only medium dependent.
For free space inside the coaxial cable, the wave impedance is always ~377 Ohm, while the line impedance is geometry-dependent and can take any possible value from almost zero to extremely large values.