One-parameter subgroups and Lie bracket Suppose that $G$ is a Lie group. It is easy to prove that given $X \in Lie(G)$ there exists a unique one-parameter subgroup $\phi_X : \mathbb{R} \to G$ such that $\dot{\phi}(0)=X$. My question is: given $X,Y \in Lie(G)$, is it possible to describe the one-parameter subgroup $\phi_{[X,Y]}$ in function of $\phi_X$ and $\phi_Y$?

Naturally, $\dot{\phi}_{[X,Y]}(0)(f) = [X,Y](f)$ where $f$ is a smooth function. Then 
$$[X,Y](f) = X(Y(f))-Y(X(f)) = X( \dot{\phi}_Y(0)(f)) - Y( \dot{\phi}_X(0)(f))$$
Or, more to the point, $\dot{\phi}_{[X,Y]}(0) = [\dot{\phi}_X(0),\dot{\phi}_Y(0)] $. More can be said if we think harder about flows along $X$ and $Y$ with respect to parameters $t$ and $s$. I'll say more a bit later.
Continuing, we either define or derive $\exp(tX) = \phi(t)$ where $\dot{\phi}(0)=X$. Hence, the question you are asking is also translated to the following question about the exponential map: how can $\exp([X,Y])$ be related to $\exp(X)$ and $\exp(Y)$ ? Of course, the rather interesting answer to this is offered by the BCH-formula:
$$ \exp(X)\exp(Y) =  \exp(X+Y+\frac{1}{2}[X,Y] + \cdots) $$
and
$$ \exp(Y)\exp(X) =  \exp(X+Y-\frac{1}{2}[X,Y] + \cdots) $$
where I used $[Y,X] = -[X,Y]$. If we could solve for $\exp( [X,Y])$ then this would give us some formula for $\phi_{[X,Y]}(1).$ Consider, 
\begin{align} \exp(X)&\exp(Y)\exp(-Y)\exp(-X) = \\
&= \exp(X+Y+\frac{1}{2}[X,Y] + \cdots)\exp(-X-Y+\frac{1}{2}[X,Y] + \cdots) \\
&= \exp([X,Y] + \cdots)
\end{align}
where I applied the BCH formula to make the last step. On the other hand, $\exp(X)\exp(Y)\exp(-Y)\exp(-X) = \exp(X)\exp(-X) = I$ thus:
$$ \exp([X,Y] + \cdots) = I$$
The next term in the $+ \cdots$ comes from
$$ \frac{1}{2}[X+Y+\frac{1}{2}[X,Y], -X-Y+\frac{1}{2}[X,Y]] = \frac{1}{4}[X+Y,[X,Y]]+\frac{1}{4}[[X,Y], -X-Y]$$
which reduces to
$$ \frac{1}{4}[X+Y,[X,Y]]+\frac{1}{4}[[X,Y], -X-Y] = \frac{1}{2}[X+Y,[X,Y]] = \frac{1}{2}[X,[X,Y]]+\frac{1}{2}[Y,[X,Y]] $$
This indicates
$$ \exp([X,Y] +\frac{1}{2}[X,[X,Y]]+\frac{1}{2}[Y,[X,Y]]+ \cdots) =  I.$$
