The Maxim MAX7221 provides constant current source and sink for every segment of every digit, set by the resistor Rset that must be connected between the Iset pin and VDD. First thing to check is if that resistor has been connected, and if so, whether it is correctly calculated for the desired current per LED.
Regardless of the number of LEDs in series, as long as the supply voltage is sufficiently higher than the forward voltage of the combination of LEDs, the current through the segment (or the dot) will be constant. In other words, the MAX7221 output pins will automatically provide different voltages for the single-LED and the double-LED current paths.
As LED intensity is related to current through it and not voltage, if any difference in intensity is observed is most likely due to difference in the optical behavior of the light-pipes used for the digit segments and the dot segment.
Adding an extra masked out LED on each of the dot connections will not expose those LEDs to 160 mA as stated in the question, but to the same current as each of the other segments is set to, via RsetM/sub>. Thus, if it provides any satisfaction, one can go ahead and add those extra LEDs in series without a problem.
Despite what another answer claims, the datasheet for the LED display clearly states:
Forward Voltage Per Segment Or (DP) 4.0 (2.0)
This means the forward voltage of the number segments is 4.0 Volts typical, while that of the dot segment is 2.0 Volts typical. Thus, the contention that the number segments are 2 LEDs each, and the dot is a single LED, is perfectly valid. It doesn't affect intensity when using constant current driving, through.
This is one of those situations where your problem is not how good you are at analysis or what base knowledge you might have, but simply that you have no clue what you don't know. This always makes the first step into electronics a very high one.
In the case of your example, what don't you know about a battery?
A great example of a larger battery with very small internal resistance is a 12 V car battery. Here, when you start the car it takes hundreds of Amps (kW of power and current in the 600 A range) to turn over the motor and the terminal voltage might drop from 13.8 V (a fully charged lead-acid car battery) to only 10 V when cranking. So the internal resistance might be (using Ohms Law) only 6 milliohms or so.
You can scale the thinking for this example to smaller batteries such as AA, AAA and C batteries and at least begin to understand the complexity of a battery.
Now what don't you know about an LED?
Now you can consider your LED. You should begin by trying to understand the datasheet for the device. While many of the characteristics you won't understand you already know one (from your question), the forward voltage (Vf) and you could probably find the current limit and maximum power dissipation in the datasheet.
Armed with those you could figure out the series resistance you need to limit the current so you don't exceed the power dissipation limit of the LED.
Kirchhoff's Voltage Law gives you a big hint that since the voltage across the LED is about 3.1 V (and the datasheet current curve tells you you could never apply 9 V), you must need another lumped model component in the circuit.
simulate this circuit – Schematic created using CircuitLab
Note: the battery internal impedance shown above is simply specified to make calculation easy. Depending on the battery type (primary or rechargeable) the internal resistance can vary. Check your battery data sheet.
Could the unknown element above simply be a piece of wire (no element)?
It could ....but we can calculate the results easily.
With two ideal voltage elements (9 V and 3.1 V) the resistors must have 5.9 V across them (Kirchhoff's voltage loop). The current flow must therefore be 5.9/10.1 = 584 mA.
The power dissipated in the LED is (3.1 * 0.584) + (0.584^2 * 10) = 5.2 Watts. Since your LED is probably rated at only 300 mW or so, you can see that it will heat up dramatically and in all probability fail within seconds.
Now if the unknown element is a simple resistor, and we want the current through the LED to be let's say 20 mA, we have enough to calculate the value.
The terminal voltage of the battery would be (9 - (0.02 * 0.1)) = 8.998 V The terminal voltage of the LED would be (3.1 + (0.02 * 10)) = 3.3 V
So the voltage across the unknown resistor is 5.698 and the current through it 20 mA. So the resistor is 5.698/0.02 = 284.9 Ohms.
Under these conditions the loop voltages balance and the LED passes its designed value of 20 mA. It's power dissipation is therefore ( (3.3 * 0.02) + (0.02^2 * 10)) = 70 mW ....hopefully well within capability of a small LED.
Hope this helps.