> Question: Suppose that finite metric space $X$ allows isometric embedding to $L_1$. Does it mean that a snowflake space $X^\alpha$ allows isometric embedding to $L_1$ for every $0 < \alpha < 1$? 

For a metric space $X = (X, d)$ and $0 < \alpha < 1$ the $\alpha$-snowflake of $X$ is a metric space $X^\alpha$ on the same set of points with the distance $d^\alpha$. 
In other words for $x,y \ \in X$ we have $d^a(x,y) = (d(x,y))^\alpha$.  

The classical [work][1] by Shoenberg provides that's this is true for $L_2$.


  [1]: https://www.ams.org/journals/tran/1938-044-03/S0002-9947-1938-1501980-0/S0002-9947-1938-1501980-0.pdf