What you really need to consider here is the batteries. The resistors are a secondary concern if you can't power the circuit. The internal resistance of your batteries is about \$187 m \Omega\$. Whatever current is going through the LEDs must also go through the batteries. The internal resistance of the battery will reduce the apparent voltage of the battery, and the power dissipated in this resistance heats the battery.
In your proposed configuration, you will have 50 parallel strings of LEDs, for a total current of \$60mA \cdot 50 = 3A\$. The internal resistance of each battery will experience a voltage drop of \$ 3A \cdot 187mA \approx 0.56V \$ and dissipate \$ 0.56V * 3A \approx 1.68W\$ of power. That power manifests as heat in the battery, and while it may be fine for short periods of time (seconds), it may overheat the battery if operated constantly. Also, you will want to minimize power in anything besides your LEDs to maximize your runtime. Every second you run this system represents 1.68 joules of energy you used to make the batteries warm instead of powering your LEDs.
It can be shown from Ohm's law that the power used in a resistor is given by \$ P = I^2R \$. We can't change \$R\$ without changing the battery, but we can minimize \$ I \$. Assuming we want the total power of the circuit to remain constant, we must raise the voltage to reduce the current, since power is the product of current and voltage, \$ P = IE \$.
If you use four batteries in series, each series LED chain can be about twice as long, since we have more voltage available. There will still be 60mA in each one, but there are half as many. So the total current required from the battery is now \$ 60mA \cdot 25 = 1.5A \$. Power losses in each battery will now be \$ (1.5A)^2 187m\Omega \approx 0.42W \$ per battery. This is still a bit much, so let's design for five batteries in series.
Five batteries will give you a voltage of \$ 5 \cdot 3.7V = 18.5V \$. We can put nine LEDs in series for a drop of \$ 9\cdot 1.55V = 13.95V \$. We want to leave more than 25% of the voltage to be dropped over the LED series resistor to get adequate current regulation. With 9 LEDs in each string, you will need 7 strings of them to get 153 LEDs. Total battery current:
\$ 60mA \cdot 7 = 420mA \$
Power and voltage lost in each battery:
\$ 420mA \cdot 187 m\Omega \approx 78mV \$
\$ (420mA)^2 \cdot 187 m\Omega \approx 33mW \$
I'll leave the calculation of the necessary resistor as an exercise. The other answers have covered that pretty well. Another interesting calculation is to calculate the total power of the circuit, the power in the LEDs, and the power in the current limiting resistors. This will give you some insight into the overall efficiency of your circuit.