Minimum speed that a bike won't fall over I thought about this when riding the bike today:
What is the minimum speed I have to go that my bike won't fall over with a given angle $\alpha$ of tilt, diameter $d$ and weight $m$ of a tire (let's say the bike tilts 30 degrees at maximum when driving in a straight line and the diameter of a tire is 0.60 m with a weight of 1 kg per tire).
I guess the angular momentum of the tires has to counter the torque of the gravitational force somehow.
How can I calculate this?
 A: The bike will not go straight if it is left alone and tipping. Secondly, it will tip slightly faster if not moving, but it will still tip over from any non-vertical angle whether moving or not.

All assumes no rider:
The right-had rule can tell you which way it’s path will curve as it tips, but your intuition will too: if leaning right the bike will gently turn right. And the handlebars/front-wheel as a unit has a lower moment of inertia around a vertical axis than the body of the bike, and hence they will turn faster.
Also, the gyroscopic “force” is just our way of trying to understand and communicate the dynamics of a spinning object when an external force is applied to its axis. The bike will tip over by itself any angle from vertical, however small, it may just take longer if moving.
Here’s one rigorous way to see the situation: the wheels are spinning and have an axis of rotation which is the same as each wheel’s axis. The direction of the vector representing the angular velocity is pointed along the axis to the left, determined from the right-had rule with your fingers going the spinning wheel’s direction, and its length determined by how fast the wheel spins.
If the bike is tipped a little, let’s say to the right, then gravity is pushing downward, and it is displaced slightly from the center of the wheel’s axis (displaced horizontally). That creates a torque around the very center of the wheel. We know a torque angularly accelerates something by $T = I \alpha , \alpha = I / T$ and the direction of this angular acceleration vector is, by the right-hand rule, pointed in the direction of the bike’s travel.
How will this change the angular velocity vector of the wheel. Just add vectors. It will transform the angular velocity vector (which points left) by moving it forward, and this means the wheel is spinning about a new axis and will move perpendicular to that new axis, ie bend to the right. The front wheels can turn more easily as they only have to move the wheel/handle-bar assembly.
If not moving, the torque will go entirely into rotating the bike itself about the point where the tires touch the pavement and not have to slowly bend the existing axis of angular momentum. If moving, the torque modifies the current dynamic by moving axis of rotation.

As an aside, the right-hand rule is an arbitrary convention; there is nothing special about right vs left. In fact, handedness is a fundamental physical parameter, which Immanuel Kant did not realize. He thought a hand in space away from any object or orientation would be neither right- nor left-handed https://philosophy.stackexchange.com/a/84452/53366
A: 
I guess the angular momentum of the tires has to counter the torque of the gravitational force somehow. Can someone help me calculate this?

That's the peculiar thing, the angular momentum of the tires doesn't actively provide torque to bring the bike back upright. Rather, it slows down the rate at which the bike can tilt, allowing a rider enough reaction time to make the proper adjustments to keep it upright.
If you get a bike up to high speed and then just "let it go", you'll notice it falls to the ground pretty quickly, almost as quickly as if you just let it go in your garage.
Edit: There's a lot of discussion below about how a bike can be self correcting due to the ability to turn - lots of great physics in there but it's missing the point of the question, which is the correction when

moving in a straight line

The amount a bike can precess to stay upright will depend on a number of parameters about the bike. On one extreme you have a wheel by itself, which will continue to turn until it slows to a stop and then fall. On the other extreme, you can imagine an extremely long bike with locked handle bars - it will fall to the ground quite quickly. All that would be a great answer to a separate question.
Moreover you'll notice I carefully worded my answer to say:

the angular momentum of the tires doesn't actively provide torque to bring the bike back upright

As I wanted to give a simple answer that doesn't overwhelm someone asking a simple question at their level (but I get that everyone loves to "actually..." on this site). This statement is correct that the angular momentum of the tires will slow the tilt toward the ground but never restore the bike toward its center. That was the point.
