What equation do you use to find these current?
You actually do need to apply Ohm's law, because the battery is not ideal and its non-ideality can be mathematically modeled as a series resistor (the battery's &quot;internal resistance&quot;). If there were no internal resistance (i.e. if the battery were an ideal 9V source) the LEDs would have failed right away from massive overcurrent.
You would have needed a multimeter (or spec sheet for the battery) to find it anyway, and it so happens that your experiment gives us enough data to find it (with some assumptions).
If we assume that the LEDs are identical and well-modeled as a constant voltage drop, we can denote \$V_f\$ the forward voltage of one LED, and \$R_{batt}\$.
The results for 3 and 4 LEDs look a bit fishy since current increased when you went to four LEDs. I'll ignore the four-LED result as an outlier for the following calculation, but recommend that you try to investigate it further.

Using the one- and two-LED circuits we can set up the following system of equations:
$$ 7\,[\text{V}] - 2V_f = 14.28\,[\text{mA}] \cdot R_{batt}$$
$$ 7\,[\text{V}] - 3V_f = 3.19\,[\text{mA}] \cdot R_{batt}$$
Solving, we get \$V_f \approx 2.1\,[\text{V}]\$, which is reasonable. However, we also find that \$R_{batt} \approx 192 \,[\Omega]\$ which is rather high. Presumably, the battery is already mostly discharged, or the currents are high enough that the simple model of a \$V = IR\$ resistor in series with the voltage source is no longer accurate due to electrochemical effects under heavy load.