I'm not an EE, but one of my goals for coming here was to be able to translate engineering-ese into layman-ese, based on the premise that if I've learned it recently, I should be able to put it into the words needed for the learner. So it may seem like I'm rambling on, but it is actually purposeful, as I am trying to paint the picture over and over again, from slightly different angles. So, here it goes...
It is my hypothesis that a simple resistor is used in most cases by most multimeters, and in one of the answers posted by Spehro Pefhany, he referenced another of his answers that gives some evidence to support my hypothesis. Just one of the multimeters he tested took the higher road and used a true constant current source.
So, going with the resistor-in-series theory...
As long as the 11 volts is held back by a suitably large resistor, the diode or LED (or LEDs) will remove their forward voltage first -- not first in terms of time, but first in terms of priority (or calculation, or mental model).
So, let's say your mystery LED is from a 60-watt equivalent LED bulb. You open it up, and your multimeter dutifully sends current through one of the LED's -- and you find that the forward voltage is 6.0 volts. This means that internally, inside the LED itself, there are actually two chips put in series, and this helps the manufacturer to accomplish their design without having to put too many 3-volt LED's on the circuit board (9 vs. 18).
So, let's assume that the resistor in series with the 6.0-Volt-LED is a 10K resistor. Then, if we do the calculations, V=IR, so I=V/R = (11V-6.0V) / 10K = 5/10,000 = 0.5mA which should be visible.
Notice how small that is. The resistor keeps the current small. Hence no blown-up LED.
Now let's test a 2N3904 NPN, the PN Base-Emitter diode junction, which we'll assume is 0.6 volts forward voltage. The calculation is (11.0V - 0.6V) / 10K = 10.4 / 10,000 or 1.04 mA. Even though the forward voltage being tested is really small, it doesn't really matter much because the resistor is comparatively much bigger.
If we just touch the test leads together and run the diode test, we get (11.0V - 0.0V) / 10K = 11.0 / 10,000 which results in 1.1mA -- the maximum current that can flow in a diode test, because there is no forward voltage taking a bite out of the total voltage range. This shows how, even though there is a total potential of 11 volts, it will never kill any LED or diode -- because 1.1mA won't kill anything.
Make your own LED tester with two 9-volt batteries in series, using a 10K resistor like we've been assuming. That's 18 volts total that we have to drop. And attach a voltmeter to the parts that go across the "diode" under test. So, the numbers testing a 3-volt white LED would be: (18V-3V)/10K = 15/10,000 = 1.5mA. And when you do this, put your volt-meter across the LED and measure the voltage, then go look up where on the V-I curve (from the datasheet for the LED) it corresponds to. Play around with different combinations of LED's and other diodes in series, and see how it works, but also do the math for a few cases, and match it to what you're measuring and seeing. If you look up LEDs on Digi-Key for instance, you will see that some LEDs have a 12-volt forward voltage. That means that, internally, such an LED has 4 chips in series (which your UNI-T may not be able to test very will). So it helps to have a decent voltage range for testing diode voltage drop -- it may be more than one diode in series (you should also be able to test some zener diodes -- 0.6 volts forward and testing backwards, something like 4.7 volts for a 1N750).
If you use a 1-ohm-resistor in series with your UNI-T meter while it is testing a 3-Volt white LED, you should be able to measure the voltage drop across the resistor to figure out how much current the UNI-T meter is actually using. This will be confirmation that there is another resistance in series with the LED, limiting the current.
Keep the 1-ohm-resistor in the chain, and track how much current is used at each step, and start adding 3-Volt white LED's in series and retesting until no current flows, then remove the last one that you just added so that current once again flows. Keep adding to the total forward voltage, now by adding RED LED's (you should need only 1, but maybe 2, in series). Then, in the same way, add garden variety 0.6 volt diodes (like 1N4148) until you've added as much forward voltage as you can, and still have current flow. In this way, you can find out if that 11 Volt maximum is real. The more LED's and other diode voltage drop you place in the test loop, the less current will flow if it is really a simple resistor in the loop. If you keep getting just about the same current flowing no matter how much total forward-voltage you insert, then you have a constant current source delivering a set amount of current, which is really the best way to implement the multimeter circuit, as it gives the widest voltage range, and also the best apples-to-apples comparison (more comparable brightness for LEDs).
Even though the diode test could be implemented using a microcontroller, with a more active, involved process, why make it so complicated, when what I've demonstrated will do the job. So that's my bet. Let us know what you find out, if you can.
Here is a constant-current-source (sink, actually) that you can try out with the two nine-volt-batteries:
Notice that in each case (short-circuited, one 3-volt-led, four 3-volt-LED's) the current flowing is about 1 mA.