User blog comment:Lord Aspect/SATCLN II/@comment-32783837-20180319184628

Q0 ≈ (10100)↑↑(10100) Q1 ≈ (10100)↑↑↑3

assuming Q(n) does the same thing to Q(n-1) as Q(1) to Q(0), which isn't specified:

Qn ≈ (10100)↑↑↑(n+2)

QX ≈ QQ(Q1+1)) ≈ (10100)↑↑↑(10100)↑↑↑(10100)↑↑↑3

QX0 ≈ Q(Q(Q1+1))+2) ≈ (10100)↑↑↑((10100)↑↑↑(10100)↑↑↑3)+2)

QXn ≈ Q(Q(Q1+1))+2(n+1)) ≈ (10100)↑↑↑((10100)↑↑↑(10100)↑↑↑3)+2(n+1))

QZ ≈ (10100)↑↑↑(3×(10100)↑↑↑(10100)↑↑↑3))

QZ0 is not properly defined, but if we assume it does the same thing as the QX sequence...

QZn = QX(QX+n+1)

QZ(QZ(QZ(QZ0))) ≈ QX(QX(QX(QX(QX))))

So the final result is hardly even distinguishable from QX, and less than your last number.

Again, what you need is a way to iterate infinitely. You'll never reach ω without that.