The code inside the Promise is not executed when you call .then, it is executed when you define the Promise itself - try just to log something at the beginning of the function:

const areallylongprocesspromise = new Promise((resolve, reject) => {
    console.log('start');
    ...
});

Since your code is synchronous, it holds the execution of the code by definition.

That happens because NodeJS and Javascript, despite allowing I/O async calls, they form single-threaded applications

It means that when the CPU is computing

    for (let i = 0; i < 1000000000; i++)
    {
        if ((i % 73829) === 0) buff++;
    }

It does not do anything else.

If you use for example setTimeout it works as expected because it creates an async timer that works as an I/O operation

const areallylongprocesspromise = new Promise((resolve, reject) => {
    let buff = 0;
    for (let i = 0; i < 1000000000; i++)
    {
        if ((i % 73829) === 0) buff++;
    }
    
    setTimeout(()=> {
       if (buff > 10) resolve(buff);
       else reject(buff);
    }, 1000)
});

areallylongprocesspromise.then((resolved) => 
{
    console.log("Resolved: ", resolved);
})
.catch((rejected) => {
    console.log("Rejected: ", rejected);
});

console.log("Waiting for execution to finish up");

This timer is useless here, just to show you how I/O async calls do work. That timer could represent for example reading a file, connecting to a DB or fetching an image from the internet, they are all I/O operations.

That's the magic of NodeJS, despite being single-threaded it does not block the code for I/O operations.