How can i increase the range of values that a light sensor gives?
The circuit is a current-to-voltage converter.  The - input to the op-amp is held at 0 (the voltage at the + input, since it is grounded) so there will be no voltage across the diodes.  The current drawn by the diodes is supplied by the op-amp output through the feedback resistor, so the larger that value, the greater the output voltage will be at any given current.  The short answer is that to increase the output voltage, you need to increase the feedback resistor. (7 ohms seems pretty small to me, unless your diodes draw a LOT of current.)
The other complicating issue is that this circuit depends on the op-amp having a low input offset voltage, since its value (~2 mV for the LM358) causes output errors.  Normally a low-offset-voltage op-amp would be used here, but maybe the use of so many diodes makes a large enough current to swamp the offset error.  If increasing the feedback R doesn't help, you might try a different op-amp (even another LM358) in case the offset of the one you have is trying to force a negative output and that's why you see 0.
Edit 11-8-20:  I looked up the sensor used in the PAR-BAR article that seems to be the source of your circuit.  It pulls about 10 microamps at 1000 lux, so 50 of them would give 500 microamps.  If you want to get (say) a 5V output at 1000 lux, you'd need a feedback resistor of 5/0.0005 = 10 kOhms.  That's without the 1.5 ohm shunt.  Also, are you applying your test illumination evenly over all 50 sensors?  The above calculation assumes that all get 1000 lux.
If you have a single sensor available, I'd recommend breadboarding with just that and a 500 kOhm feedback resistor, and no shunt.  That way there will be no issue about equal illumination.  If you still have problems, and you are committed to the LM358 for some reason, try using a simple circuit to compensate for input offset.  These typically involve a potentiometer (could be a trimmer in the final unit) that adjusts a very small voltage applied to the + input.  In your tests you may find that it needs to be negative, which you would have to apply from some external source (since there is no - supply on the Arduino).  If that's needed to resolve your problem, consider a better op-amp or a different circuit configuration.
Edit 11-9-20: Had another look at the PAR-bar article and noticed that they are measuring differential voltage across the photodiode array, which is not the same as your current-to-voltage converter (although that is indeed typical of photodiode circuits).  The voltages that they get are 30 mV or less.  I suggest you get a DMM and measure the voltage across the array, without any connection to other circuitry.  Cheap DMMs have 200 mV full-scale, which should at least show you what's going on.  The article used an expensive datalogger and only suggests that Arduino-type solutions may be possible, but gives no circuits.  You probably don't need a true differential amp, since the photo voltage is always in the same direction.