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#Python 3, f_ω^ω6(f_ω^ω5 (9e999)) (non-standard fundamental sequences)

exec('''def a(n):
 if n:
  a(n-1)
  c="a%s="%(n+1)
  d="a%s(a%s)("%(n,n+1)
  for i in range(n+1):
   c+="a4=`a%s:"%(n-i)
   if i-n:d+="a%s)("%(n-i)
   exec(c+d+"a0-1))if x else 9")
g=`n,x:exec("a(x)")or(eval("a%s"%n)if n-x else g(n+1,x)(eval("a%s"%n)))
b=`f:`x:g(2,x)(f)(x)
c=`b:`f:`x:g(3,x)(b)(f)(x)
d=`c:`b:`f:`x:g(4,x)(c)(b)(f)(x)
e=`d:`c:`b:`f:`x:g(5,x)(d)(c)(b)(f)(x)
h=`e:`d:`c:`b:`f:`x:g(6,x)(e)(d)(c)(d)(f)(x)
print(g(8,h(e)(d)(c)(b)(`x:x+x)(9e999))(h)(e)(d)(c)(b)(`x:x+x)(99))'''.replace('`','lambda '))

Try it online! Unfinished ungolfed version! (same method)

the a(n) function defines the series of functions a2,a3,a4,…,aN,aN+1

#X→Y I will use this to illustrate the domain and range of a function,
#X→Y means the function takes an input of type X and an output of type Y
#(N is any integer)

a2(f)(x)=[a2(f)](x)~=f^x(x)       ---a2 itterates f,(N→N)→(N→N) ,
                                  ---a2 generates a more powerful f,
                                  ---   +1 (in the fast growing hierachy the ordinal increases by 1 every time the function is used)
a3(a2)(f)(x)~=a2^x(f)(x)          ---a3 itterates a2, [(N→N)→(N→N)]→[(N→N)→(N→N)], 
                                  ---   +ω 
a4(a3)(a2)(f)(x)~=a3^x(a2)(f)(x)  ---   +ω²
a5                                ---   +ω³
aN                                ---   +ω^(n-2)

b(f)(x)=aX(aX-1)(aX-2)…(a3)(a2)(f)(x)   ---an a2 type function (N→N)→(N→N)
                                  ---but using the full power of the a-series,
                                  ---   +ω^ω
c(b)(f)(x)=aX(aX-1)(aX-2)…(a4)(a3)(b)(f)(x)   ---an a3 type function
                                  ---   +ω^ω² (I think)
d(c)(b)(f)(x)=aX(aX-1)(aX-2)…(a5)(a4)(c)(b)(f)(x)
                                  ---   +ω^ω³ (I think)
e                                 ---   +ω^ω^4 (I think)
h                                 ---   +ω^ω^5 (I think)
g(8,x)(h)(e)(d)(c)(b)(f)(x)       ---   +ω^ω^6 (I think)

the base function in my code is x+x=f_1(x) but it doesn't really matter since it is consumed by +ω^ω^6 so my function is at the f_ω^ω^6 level.
The first input is much more important (since it determines "the collapse") and it is value is h(e)(d)(c)(b)(x:x+x)(9e999)which bigger thanf_ω^ω^5(9e999)` (if I am correct).

So the final number is f_ω^ω^6(f_ω^ω^5(9e999)) with non standard fundamental sequences.