Let $j$ be an integer.
Does there exist an elliptic curve $E_j$ over $\mathbf{Q}$ with $j$-invariant equal to $j$ whose minimal discriminant we can write down in a practical way?
For example, can we write down an elliptic curve $E_j$ as above for which we know the reduction over $\mathbf{Z}$ and we know a number field $K$ such that $E_j$ obtains semi-stable reduction over $K$.
If by "practical way" you mean a fast algorithm, I think that's possible. In $char \neq 2,3$ $$j=1728 \frac{c_4^3}{c_4^3-c_6^2}$$ (see http://en.wikipedia.org/wiki/J-invariant#Algebraic_definition ) Therefore given a $j$-invariant we can solve for $c_4$ and $c_6$, and if they can be chosen to be rational, we already have a curve $E$ over $\mathbb Q$ to start with. I assume for convenience that $j \neq 0,1728$.
Given $E$, we can use the process Stein and Watkins applied to construct their table, see http://modular.math.washington.edu/papers/stein-watkins/
The relevant part is
Here $E_\tilde{p}$ refers to the quadratic twist by $\tilde{p}$. It appears to me you could twist by $\tilde{p}$ and repeat this process until you arrive at a twist of minimal conductor (apart from $2,3$), then you can find its minimal discriminant.
For $2$ and $3$ there are possibly more complications, but you can just do trial and error among the finitely many possible twists.
p.s. When $E$ has nontrivial $p$-torsion over a number field $K$, reduction at all places of $K$ not lying over $p$ must be semistable. (e.g. see the paper "Abelian varieties having purely additive reduction") You can construct many fields this way where the curve becomes semistable.