Take the iterated function one iteration at a time.

f(x) = x^2 + 12x + 35

minimum:  2x+12=0; x = -6; f(-6) = -1

(-6, -1) for one iteration

f of f:

(x^2 + 12x + 35)^2 + 12(x^2 + 12x + 35) + 35

x^4 + 144x^2 + 1225 + 24x^3 + 70x^2 + 840x + 12x^2 + 144x + 420 + 35

x^4 + 24x^3 + (144+70+12)x^2 + ( 840+144)x + (1225 + 420 + 35)

f(f(x)) = x^4 + 24x^3 + 226x^2 + 984x + 1680

minimum:

4x^3 + 72x^2 + 452x + 984 = 0

x^3 + 18x^2 + 113x + 246 = 0

(x+6)(x^2 + 12x + 41) = 0

x = -6  plug into:

x^4 + 24x^3 + 226x^2 + 984x + 1680

1296 - 5184 + 8136 -  5904 + 1680 = 24

(-6, 24)  for two iterations

f of f of f:

(x^2 + 12x + 35)^4 + 24(x^2 + 12x + 35)^3 + 226(x^2 + 12x + 35)^2 + 984(x^2 + 12x + 35) + 1680

At this point I gave up and started using Wolfram Alpha to expand each term, one at a time.

x^8 + 48 x^7 + 1004 x^6 + 11952 x^5 + 88566 x^4 + 418320 x^3 + 1229900 x^2 + 2058000 x + 1500625

+ 24 x^6 + 864 x^5 + 12888 x^4 + 101952 x^3 + 451080 x^2 + 1058400 x + 1029000

+ 226 x^4 + 5424 x^3 + 48364 x^2 + 189840 x + 276850

+ 984 x^2 + 11808 x + 34440

+ 1680

f(f(f(x))) = x^8 + 48x^7 + 1028x^6 + 12816x^5 + 101680x^4 + 525696x^3 + 1730328x^2 + 3318048x + 2842595

I'm going to guess that the minimum occurs at x = -6  (so my solution is not fully valid)

Plugging in -6 yields:   899

(-6, 899) for three iterations

I do not believe that, in general, a series of iterated functions always have their minimum at the same x value, but since it was true in this case for f(x) and for f(f(x)) I'm guessing it is true for f(f(f(x))) in this case.  This still may not be true for all iterations.  In this case Desmos confirms.

https://www.desmos.com/calculator/9hrefnw6y2

There is probably an easier algebraic method, but if so, it escapes me.