Why is the running cost of SQUARE-MATRIX-MULTIPLY-RECURSIVE in CLRS $\Theta(n^3)$ instead of $\Theta(n^2)$?
What you are asking is indeed answered in section 4.5 of CLRS using a methodology described later in the book. I will post here an alternative answer you should be able to understand with your current knowledge. 

Think of the recursive relationship
 $$T(n) = 8T(\dfrac{n}{2}) + cn^2.$$

Let us first see what happens if we substitute $T(\dfrac{n}{2})$  

$$T(n) = 8 (8T(\dfrac{n}{4}) + c (\dfrac{n}{2})^2) + cn^2 = 8^2 T(\dfrac{n}{4}) + 8c(\dfrac{n}{2})^2 + cn^2.$$

By substituting $T(\dfrac{n}{4})$ we get:

$$T(n) = 8^2 (8T(\dfrac{n}{8}) + c (\dfrac{n}{4})^2) + 8c(\dfrac{n}{2})^2 + cn^2 = 8^3 T(\dfrac{n}{8}) + 4cn^2 + 2cn^2 + cn^2.$$

It becomes obvious that in the general case after $k$ iterations we have:

$$T(n) = 8^k T(\dfrac{n}{2^k}) + cn^2(2^{k-1} + 2^{k-2} + \dots + 2 + 1).$$

Let $n = 2^k$. 

Notice that $k = \log_2n \Rightarrow8^k = 8^{\log_2n} = 2^{3\log_2n} = 2^{\log_2n^3} = n^3$. By substituting $k = \log_2n$ in the previous relationship we get:

$$T(n) = n^3T(1) + cn^2 (\dfrac{1}{2} + \dfrac{1}{4} + \dots + \dfrac{1}{n})n.$$

$T(n) = n^3 T(1) + cn^3$, therefore you have $\Theta (n^3)$.