Solution:

If every vertex is $1$, we get $14$ and that is clearly the highest possible total. The lowest possible total cannot be lower than $-14$, but we cannot even achieve that because if all the vertices are $-1$, then all the faces are $1$.

If we change a vertex, then we also change three faces. If the vertex and the three faces are all initially the same, then we make a change of $\pm 8$. If three are of one kind and one the opposite, then we make a change of $\pm 4$. If two are of one kind and two the opposite, then we make no change. Thus any sequence of changes must take us to $14 + 4n$ for some integer $n$. But we have already shown that the total is greater than $-14$ and at most $14$, so the only possibilities are $-10, -6, -2, 2, 6, 10$ and $14$.

We show that $10$ is not possible. If more than $2$ vertices are $-1$, then the vertex total is at most $2$, there are only $6$ faces, so the total is less than $10$. If all vertices are $1$, then the total is $14$. If all but one vertex is $1$, then the total is $6$. So the only possibility for $10$ is just two vertices $-1$. But however we choose any two vertices, there is always a face containing only one of them, so at least one face is $-1$, so the face total is at most $4$ and the vertex total is $4$, so the total is less than $10$. The other totals are possible, for example:

$14$: all vertices $1$

$6$: one vertex $-1$, rest $1$

$2$: three vertices of one face $-1$, rest $1$

$-2$: all vertices $-1$

$-6$: all vertices but one $-1$

$-10$: two opposite corners $1$, rest $-1$