Is it possible to build a computer that would output $10^{10^{100}}$ symbols and halt, without using ~$10^{100}$ space?
The answer really depends on your computation model. On a (fixed) Turing machine, you can count up to $n$ using no space. A better formulation of the question is like this:


  Is there a function $f\colon \mathbb{N} \to \{0,1\}^*$ satisfying $|f(n)| = o(\log n)$ and a program $P$ such that $P(f(n))$ uses space $O(f(n))$ and outputs $1^n$?


It is a basic result of Kolmogorov complexity that this is impossible, even without the restriction on space. However, if you are willing to allow randomness and to incur some (multiplicative) inaccuracy, you can use the approximate counting technique of Morris and Flajolet, for which $|f(n)| = \Theta(\log \log n)$.