From which dimensionful constants does proton mass arise? It is well known that the most of the proton (or any other hadron with light quarks) mass is not made up from quark masses, but it is dynamically generated by QCD mess inside. I've also heard that, even if quarks would be massless, protons (and other hadrons) would still have a nonzero mass.
However, if proton mass does not (in most part) arise from the quark masses, from which dimensionful constants does it arise?
I've heard that proton mass arises from spontaneous symmetry breaking of scale invariance. However, this is a troublesome explanation, or non-explanation at best, because it opens more questions:
If a theory is scale invariant, how can it pick a scale when breaking this symmetry? The proton mass is a constant, so how can the scale invariance be broken across entire universe the same way? Is there a field, very resistant to change, that permeates entire space to ensure the constancy of proton mass?
 A: I think its easiest to understand this if one has a minimal understanding of QFT. I'm not sure about your background knowledge but hopefully this isn't gibberish to you.
The QCD Lagrangian for massless quarks is given by,
\begin{equation}
{\cal L} = - g \sum_i \bar{\psi} _i  A _\mu \gamma ^\mu \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu } 
\end{equation}
where the fields are, $ A _\mu $ and $  \psi _i $. The only constant in the equation is the coupling constant, $g$. Therefore, we see that there is no single scale in the Lagrangian. Naively one would say that the theory is scale invariant.
However, there is a subtlety. We haven't fully specified the theory. We have yet to say what the value of the coupling constant is. The problem is that QFT causes the strength of an interaction to depend on the scale which its measured. Luckily, we know how to calculate how a coupling changes with scale (this is done in every full year QFT course),
\begin{equation} 
\frac{ d \alpha }{ d \log \mu } =  - \frac{ b  }{ 2\pi }  \alpha ^2 
\end{equation}
where, $ \alpha \equiv g ^2/4 \pi  $ and $ b  $ are calculable numbers. For QCD with the SM fermions we have,
\begin{equation} 
b  = 7 
\end{equation}
From here its easy to solve the differential equation above and get the coupling as a function of the scale, $ \mu $,
\begin{align} 
\frac{1}{ \alpha ( \mu ) } &= \frac{1}{ \alpha ( \mu _0 ) } - \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 }  \\
\alpha_s (\mu) &= \frac{ \alpha _s ( \mu _0 ) }{ 1 + \alpha _s ( \mu _0 ) \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 } }
\end{align}
Therefore, we can measure the coupling at some scale and then know what it is at every scale. As pointed out by the OP, we can already see breaking of scale invariance since the couplings depend on scale.
Now we move on to the relation to $ \Lambda_{QCD} $. This is conventionally defined as the scale where the coupling becomes infinite. From the running above we see this occurs when,
\begin{equation} 
\mu \equiv \Lambda_{QCD} = \mu _0\exp \left[ - \frac{ 2\pi }{ b  \alpha _s ( \mu _0 ) } \right]
\end{equation}
Here we see that the scale only depends the field content (through $b$) and Natures choice for the coupling.
