If you look at the data sheet, the secondary DC resistance is 0.6 ohms. The recommended load resistance is 50 ohms. As the data sheet says, this value was selected to give a 1 volt output for 1 ampere flowing through the primary. As a check, 1 ampere in the primary will yield (for a 50:1 ratio) 20 ma in the secondary. With 20 ma through it, the 50 ohm load resistor will indeed produce 1 volt of output. With that said, the minimum current that can be sensed is probably limited by the noise level of your sensing circuit. With 10 ma in the primary the secondary level will be 0.2 ma which will yield 10 millivolts across the 50 ohm load resistor. This should be well above any electronic noise in your sensing circuit.