How to conect multiple output pin from sensors to a single digital pin on a arduino
If the ICs output 5V and your Arduino is 3.3V, you can simply use one voltage divider at the Arduinos pin to divide the voltage down to 3.3V.

But there is still one important thing to connect them safely: You have to check, if the outputs of the ICs are driven actively high, when no fault occured, or if they simply use a pullup resistor internally.

A pin, that is actively driven high, does not have an overcurrent protection. So, if one IC would pull it's pin to low and the others are still high, a high current would flow between them, potentially destroying them. In that case you need current limiting resistors in series with the ICs (directly at the ICs, so that the resistors are between them).

On the other side, if the ICs output a high level through a pullup resistor, you are safe to connect them directly, as there can only flow a very little current through the pullup resistors. Often such a situation is called Open-Drain (passive high level through pullup resistors, active low level by driving the pin to low through a MOSFET, like used in a digital output pin). In that case you don't need the current limiting resistors.



EDIT: The datasheet of the ICs, that you linked to states that the fault output is an


  Active-low open-drain output


Also the example circuit shows an external pullup resistor. That means, that the fault pin is internally connected to a MOSFET, which will connect the pin to ground in case of a fault. Otherwise the pin is floating. The external pullup resistor will pull the pin state to a high level, when the MOSFET is not activated. This is good for you, since you can now simply use 1 resistor (one resistor is enough for all ICs), to pull the level up to 3.3V. Then you directly have the correct level for your Arduino and you don't need a voltage divider.