Using LEDs in a bridge rectifier circuit
LEDs typically have much lower reverse breakdown voltage limits (Vrrm = 5 to 15 Volts) than even the cheapest, smallest silicon diodes (1n4001 = 50 Volts , 1n4007 = 1000 Volts, 1n4148 = 100 Volts). 

 (Wikipedia)

Forward voltage Vf on the other hand is anywhere from 1.7 Volts for some red LEDs, to 3.5 Volts or higher for some blue and white ones. Compare this to a typical 0.7 Volt Vf for a standard silicon rectification diode.

In a full bridge rectifier configuration, the voltage is dropped by 2 x Vf during the forward conduction part of the cycle. 

Thus if you were to build a bridge rectifier with LEDs, the output voltage would drop anywhere from 3.4 to 7 Volts or more, compared to the input. The conduction would start much later into the positive part of the cycle (once the voltage rises above the LED Vf), compared to a silicon diode. It would also end earlier. Also, during the reverse part of the cycle, depending on the specific LED used, the "diode" is likely to enter conduction well within the input voltage range. 

In other words, the bridge will provide voltage during a much smaller part of the AC cycle than with silicon diodes, and will provide a lower voltage even then.

As your purpose seems to be actually lighting the LEDs rather than just rectifying the voltage, the output of the bridge may not matter, but it's useful to know.

A suggested alternative if you are keen to use LEDs in this particular way, is to hook up the LEDs as you desire, but to add one inexpensive and tiny 1n4001 or similar, in series the current limit resistor and each LED. The diodes are not much bigger than the discrete resistors you will probably use. 

Why this works:
The silicon diodes will block reverse voltage better than an LED can, i.e. to a much higher voltage.

However, the recommendation remains to use just an integrated 4-lead bridge IC (35 cents single unit, 800 Volts/1 Ampere), and use the LEDs as in any DC circuit configuration. Space requirement will not go up significantly.



Regarding current limiting resistor calculation:

Obtain the peak voltage of the bicycle dynamo's AC signal, using a multimeter with either peak AC voltage mode if available, or RMS AC voltage (more common in basic multimeters) and multiplying by 1.4142 to estimate the peak voltage. All this while pedalling as fast as you can with the cycle on its stand.

Add in a safety factor by doubling this peak voltage if you are concerned about blown LEDs due to high voltage when speeding downhill, for instance. LEDs are somewhat forgiving of minor current spikes if they are for brief durations - It isn't over-current that kills an LED usually, it is inability to get rid of generated heat.

Now, if Vmax is your safety-adjusted peak voltage, subtract your LED Vf from it (2 x Vf if you are going with the bridge, 2 x (Vf + 0.7) if you add in the silicon diodes), to obtain the voltage Vres each resistor needs to drop.

R = Vres / I, 

So plug in the recommended current I for your LED of choice, and you have your numbers. 

LEDs commonly come in 5 mA (SMD and some 3mm ones), 20 mA, 25 mA or in the region of amperes (e.g. Osram "Golden Dragon" LEDs). Refer to appropriate datasheets.