Suppose for the moment that $n=2$. Then the birational map $f\colon \mathbb{P}^n\dashrightarrow(\mathbb{P}^1)^n$ described in the question, which is $f\colon [x:y:z] \dashrightarrow([x:y],[x:z])$ has two base-points, $[0:0:1]$, $[0:1:0]$ and the inverse is $f^{-1}\colon ([a:b],[c:d]) \dashrightarrow[ac:bc:ad]$ and has one base-point, namely $([0:1],[0:1])$. If $X\to \mathbb{P}^2$ is the blow-up of $[0:0:1]$, $[0:1:0]$ and $X'$ is the blow-up of $(\mathbb{P}^1)^2$ at $([0:1],[0:1])$, the corresponding map $X \dashrightarrow X'$ is an isomorphism, and thus a "crepant" birational map. If you want to blow-up more points on $X$ and $X'$, which corresponds via this isomorphism you again get an isomorphism. If you blow-up other points you get a birational map which is not an isomorphism and thus not crepant (between two smooth projective surfaces crepant=isomorphism).
For $n>2$, the birational map $f$ has more that only isolated base-points, so you need to blow-up more complicated stuff, then contract, and maybe blow-up again... and can get a sequence of simple birational maps (blow-ups or inverse of blow-ups) from $\mathbb{P}^n$ to $(\mathbb{P}^1)^n$. In any case, you cannot get morphisms if you only blow-up points of $\mathbb{P}^n$.