So I think it's true that a field $F$ and its respective vector space $F/F$ are isomorphic, since they consist of the same elements, and the operations of $F$ (addition,mult.) and $F/F$ (vector addition, scalar mult.) are essentially the same.
But what if I am talking about a NLS $F/F$? Then we have a topological aspect to account for. By scalar property of a norm we have $\|\alpha v\| = |\alpha|\|v\|$. But isn't absolute value a generalized notion (https://en.wikipedia.org/wiki/Absolute_value_(algebra))? Then isn't norm dependent on which absolute value function we are talking about?
Supposing we choose a specific absolute value function in regard to our norm, what can we say about $F/F$? I think I found that any two norms are then isometric, hence that there is only "one" norm, that is, the one induced by $|\cdot|$. Then can we essentially view $F$ and any NLS $F/F$ as the same mathematical structure?
I assume NLS is an abbreviation of normed linear space? To be precise: let $F$ be a field and let $|\cdot|$ be an absolute value function on $F$. A normed linear space over $(F,|\cdot|)$ is a pair $(V,\|\cdot\|)$ consisting of a vector space $V$ over $F$ and a norm function on $V$.
So, as you suspected, the norm "depends on the absolute value".
For a norm on $F$ one has $\|1\|=\|1\cdot 1\|=1\|1\|$ and therefore $\|1\|=1$. Hence $\|v\|=\|v\cdot 1\|=|v|\cdot 1=|v|$. This shows that there exists only one norm on $F$, namely the given absolute value.
So the answer to your last question is "yes".