Insert a 100 ohm or so resistor in series with the LED, to limit the current if you get the transistor fully turned on.
Disconnect the audio signal for now.
Instead of your BE and BC resistors, use a pair of resistors, one from B to +5, and the other from B to ground. The ratio should bias the base so that the LED is just dark. That should be about 10k to +5 and 1k to ground -- but adjust until the LED is slightly off.
4 Connect the audio to the base via a capacitor -- this is actually the crucial ingredient. A range of values will probably work -- say 10uF. That will be polarized -- the + end should go to the base.
That should be much more sensitive than what you've got. Your main issue is that the audio is likely at zero volts, and so you need about 0.5V to 0.7V peaks (on the positive side) to get the transistor to turn on. Using biasing resistors and AC coupling with the cap means you need much smaller +ve peaks to get some LED lighting.
You cannot make both assumptions about \$V_{ce} = 0.2V\$ and \$\beta = 100\$ at the same time; they are for two different modes of operation for the transistor (there's also a third mode where the transistor is in cutoff, and a fourth less common one in reverse active).
You start out assuming one of the possible states. For example, suppose I assume the transistor is saturated. Then \$V_{ce} = 0.2V\$ and \$V_{be} = 0.7V\$. You can then solve for base and collector currents using these assumptions for the BJT and only these assumptions.
The last step is to check your assumptions. For example, to make sure the transistor isn't actually in the linear active region must check that \$\beta I_{b} \gg I_{c}\$. So let's say in your case you get \$I_{c} = 9.8 mA\$ and \$I_{b} = 9.8 \mu A\$. Well clearly our assumption about saturation was bad because \$\beta I_{b} = 980 \mu A\$, which is less than our calculated \$I_{c}\$. We must then start over with new assumptions and re-solve the problem.
Instead, suppose \$I_b = 2 mA\$. Then \$\beta I_b = 200mA\$, which is much greater than \$I_c\$, so now the saturation assumption is correct.
Note: \$\beta I_b \gg I_c\$ is kind of a vague limit. Typically we use at least 10 times bigger for saturation.
For solving assuming the BJT is forward active, you would assume \$\beta I_b = I_c\$ and only this assumption. To check it (against saturation), simply make sure \$0.2V < V_{ce}\$.
Best Answer
What you want to do is saturate the transistor, which means you want a base current (very approximately) 0.1* the collector current - around 0.3mA, or higher currents won't hurt. Assuming you're switching from a 3.3V MCU output, you'll have about 2.6V across the base resistor after subtracting Vbe, so Rb should be less than 2.6/0.3 = 8.67kilohms (choose 8k2 or even reduce to 4k7).
What voltage does that give between collector and emitter (Vce)? Check the datasheet for your chosen transistor; it'll probably guarantee "below 0.2V". e.g. for the BC847, Table 7 shows "VCEsat : IC=10mA; IB=0.5mA 90(typ) 200(mak) mV" So Vce would be under 0.1V at currents below 10mA - and note the base current is only 0.05* the collector current.
Also see the graph in Figure 3 which shows measured performance of a sample transistor, where Vce (at 25C) is only around 50mv under these conditions, rising to 100mv at 30mA.
If that's not enough, the higher gain BC847 reduces Vce to about 30mv under the same conditions (see Fig.11)
Most small signal NPN transistors should have similar info in the datasheets, I've just used the BC847 (aka BC107 for old timers) as an example.