Let's denote the common ratio of the geometric sequence as $q$. We also know that $a_1=1$ and $a_4=a_1q^3=8$.

Solving for $q$ gives us:
$$8 = 1 \cdot q^3$$
Therefore, $q = 2$.

Now, we can find $S_5$ using the formula for the sum of the first $n$ terms of a geometric sequence:
$$S_5 = \frac{a_1 \cdot (q^5 - 1)}{q - 1}$$
Substituting the values of $a_1$ and $q$ into the formula, we get:
$$S_5 = \frac{1 \cdot (2^5 - 1)}{2 - 1} = \frac{32 - 1}{1} = 31$$

So, the sum of the first five terms of the sequence is $\boxed{31}$.

This solution demonstrates the application of the general formula for the sum of a geometric series and shows knowledge of the formula for the nth term of a geometric series. It is a basic problem type that tests understanding of geometric sequences.