A similar question has already been asked on dwarf planet Ceres: Could a Human reach escape velocity by jumping from the surface of Ceres (a dwarf planet)?
Ceres has 2.9% of Earth's gravity. Saturn's moon Mimas has 0.6% of Earth's gravity. If you jump strong enough, could you fly off into outer space from Mimas?
tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_{esc} = \sqrt{\left(\frac{2 GM}{r_0} \right)}, $$
showing that it is the $\frac{mass}{radius}$ ratio that's key here, not just the surface gravity given by
$$a_{g} = -\frac{GM}{r_0^2}. $$
So since
$$v_{esc} = \sqrt{a_g r_0}, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = \sqrt{2 g h}$ and $t = \sqrt{2x/g}$.
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, the surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are 160 and 510 m/s, respectively!
So... no chance, not even close!
No.
Fermi estimate: Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
When thinking about escaping a moon most people use the moon's 2 body escape velocity, sqrt(2GM/r). However reaching the edge of the moon's Hill Sphere can be sufficient.
The Saturn Mimas L1 (SML1) is at an altitude of about 332 kilometers.
In a 2 body scenario it would take .14 km/s to get an apo-apsis altitude of 332 kilometers. However tidal forces tend to boost the velocity on the way to EML1 so it might take somewhat less.
.14 km/s is about 300 mph. So no, not doable by jumping.
Tidal influence is especially pronounced on Phobos. Mars-Phobos L1 and L2 are only 3 or 4 kilometers from the ends of Phobos. the boulders in the vicinity of Stickney Crater are already nearly weightless from tidal influence and the same is true for the other end of Phobos. If Phobos goes just a little lower it will go beyond the Roche Limit and this moon would be smeared into a ring.
So I expect it would take a lot less than what we call bicycle speed to escape from either Deimos or Phobos.
However a bike's acceleration is proportional to force of friction on the tires. Which is proportional to net acceleration towards the Martian moon. So I don't think Randall Munroe's bicycle would do much more than spin its wheels.
