Amplitude of Random Cubic Graphs
The amplitude $\alpha(n) = \Theta(n)$  for expander graphs. A random 3-regular graph is asymptotically almost surely an expander graph (see Wikipedia), so the expectation of the amplitude will be $\Theta(n)$, since the probability that it's not an expander graph goes to $0$ as $n$ goes to $\infty$.

For an expander graph with parameter $\beta$, for any set of $s$ vertices with $s \leq n/2$, there are $\beta s$ neighbors of the set. Now, let the number of vertices on level $j$ be $\ell_j$, with $\ell_0=1$. We then have from the expansion property that as long as $j$ is not too large (i.e., we haven't included half the vertices yet)
$$
\ell_j \geq \beta\, \sum_{i=0}^{j-1} \ell_{i} 
$$
Now, look for the level $\ell_j$ which contains vertex $\frac{n}{3}$. That is, so $\sum_{i=0}^{j-1} \ell_i &lt; n/3$ and $\sum_{i=0}^{j} \ell_i \geq n/3$. If this level is large, i.e., $\ell_j \geq n/6$, we are done. Otherwise, the next level has size 
$$
\ell_{j+1} \geq \beta \, \sum_{i=0}^{j} \ell_{i} \geq \beta \frac{n}{3},
$$
and we are done. 

While this proof looks at the number of vertices in a level rather than the number of edges (which the OP asked about), there are always at least as many edges added in step $i$ as vertices in level $i$, since each vertex must be reached by some edge.