First, Field strength.
This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz.*
For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m.** This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects:
1) The ~1/r variation in the electric field (reduction to 3 kV/m). I used the method of images to calculate this field, with a 10 cm conductor diameter to keep the peak field below the 1MV/m breakdown field.
2) Cancellation from the other two power lines in this 3-phase system, which are at +/-120 degree electrical phases with respect to the first, and are physically offset in a horizontal line per the photo. I estimated 7m spacings between adjacent lines. The maximum E-field actually occurs roughly twice as far out as the outermost line; the field under the center conductor is lower.
Next, Can you feel it?
1) The human body circuit model for electrostatic discharge is 100pF+1.5kohm; that's a gross simplification but better than nothing. If one imagined a 2m high network, the applied voltage results in a 50Hz current of about 70uA ($C \omega V$). Very small.
2) There will be an AC voltage difference between the (insulated) human and (insulated) bicycle. A 1m vertical separation between their centers of gravity would yield roughly 1200V. This voltage is rather small compared to some car-door-type static discharges, but it would still be sufficient to break down a short air gap (but not a couple cm), and would repeat at 100Hz. I imagine it would be noticeable in a sensitive part of the anatomy.
If the transmission voltage is actually 400 kV, all the field strengths and voltages would of course double.
(*) In response to a comment, here's an estimate of the neglected induction and radiation effects, courtesy of Maxwell 4 and 3:
Induction: Suppose a power line is carrying a healthy 1000A AC current (f=50 Hz). Then by Ampere's law, there is a circumferential AC magnetic field; at the wire-to-ground distance of 20 meters that field's amplitude is $10 \mu T$. (Compare with the earth's DC field of approximately 0.5 gauss, or $50 \mu T$.)
The flux of this magnetic field through a $1 m^2$ area loop (with normal parallel to the ground and perpendicular to the wire) is $\Phi = 10 \mu Wb$ AC. Then from Faraday's law, the voltage around the loop is $d \Phi /dt = 2 \pi f \Phi = 3 mV$ (millivolts). So much for induction.
One can also estimate the magnetic field resulting from the $1200 V/m$ ground-level AC electric field, which has an electric flux density $D =\epsilon_0 E = 10.6 nC/m^2$ and a displacement current density $\partial D / \partial t = 2 \pi f D = 3.3 \mu A/m^2$. The flux of this field through a $1 m$ square loop (parallel to the ground) is $3.3 \mu A$, so the average magnetic field around the square is $0.8 \mu A/m$, for a ridiculously small magnetic flux density of $1 pT$.
(**) 1 Sep 2014 update. Dmytry very astutely points out in a comment that there will be local electric field intensification effects from conductive irregularities in the otherwise flat ground surface, such as our cyclist (who, being somewhat sweaty, will have a conductive surface). The same principle applies to lightning rods.
For the proverbial spherical cyclist, the local field will be increased by a factor of 3, independent of the sphere's size, as long as it's much less than the distance to the power line. It turns out that it doesn't matter whether the sphere is grounded or insulated, since its total charge remains 0.
For more elongated shapes the intensification can be much higher: for a grounded prolate spheroid with 10:1 dimensions, the multiplication factor is 50. This intensification of course enhances any sensation one might feel.
If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground.
To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power line has a resistance $R$, the voltage across the line is $V_w=IR$, which is smaller for a smaller current. Thus, to supply a fixed amount of power to the load, if the voltage at the load $V_L$ is higher, the voltage drop across the power line $V_w$ is smaller, so the power loss in the power line, $V_w^2/R$, is smaller.
Putting all those pieces together, the power loss in the power line is
$$P_w=\frac{V_w^2}{R}=\frac{(IR)^2}{R}=I^{2}R=\left(\frac{P_L}{V_L}\right)^2 R\ .$$
Best Answer
I would add, however, that any modification of the environment near the transmission lines changes the net impedance (the effective AC resistance) of the circuit carrying the power. So, even while for the most part you are just collecting leaked energy, you are also lowering the amount of transmitted energy. If you build an electric circuit containing a coil and this will deliver the a certain amount of energy to all components. You can then bring a second coil near the first, without touching it, and the currents, potential drops and power in the elements will change.