Why does Lagrangian of free particle depend on the square of the velocity ? Why does Lagrangian of free particle depend on the square of the velocity ? 
For example, $L(v^4)$ also doesn't depend on direction of $v$. 
 A: The Lagrangian should not only be independent of the direction of $\vec{v}$ but it should also change correctly under a Galilean transformation. For instance, if $K$ and $K'$ are two frames of reference with a relative velocity $\vec{V}$ then the two Lagrangians $L$ and $L'$ should differ only by a total time derivative. If $L$ is a function of fourth power of velocity then $v'^4 = (\vec{v} + \vec{V})^4 = v^4 + \{V^4 + (\dot{\vec{x}}\cdot\vec{V})^2 + 2v^2V^2 + 2(\dot{\vec{x}}\cdot\vec{V})v^2 + 2(\dot{\vec{x}}\cdot\vec{V})V^2\}$. The term in the curly brackets cannot be expressed as a total time derivative. If $L$ was $v^2$, it would have been possible.
Galilean invariance forces Lagrangian to be a quadratic function of velocity. You may want to read section 4 of Landau and Lifshitz's Mechanics to understand this point better.
A: I think that the reason is merely a matter of convenience to use |v|^2 instead of e.g. |v|.
Not until later in the text they prove that The Lagrangian is linearly proportional to |v|^2 
