There are a lot of high-quality answers here and each one of them replies to one or more specific part of the original question, but unfortunately no answer addresses all the question.
Since I don't know which one to accept I decided to post this Community-wiki answer that summarizes all the (pre)existing ones.
Can you draw a polygon with:
- 6 sides and area 6?
- 8 sides and area 8?
- 12 sides and area 12?
Example of solutions from @Glorfindel's answer:

What about a polygon with 13 sides and area 13?
From @Athin's answer (but also @Glorfindel's):
No [it is not possible to construct the requested polygons for each $n$]. Trivially, n≤2 is impossible. For odd n, it's also impossible because the polygon must have an even number of sides (the sides alternately change from horizontal and vertical.) So we are only dealing with even n≥4.
Can you find a constructive approach to draw any polygon with $n$ sides and area $n$?
Again from @Athin's answer:
Sure thing! Are you interested in worms btw? :) 
Bonus question: As you can easily see there are many solutions for each $n$. Among all polygons that fulfill the requirements for a fixed $n$, which one is the "smallest" one? i.e. which one can be inscribed in the rectangle with the lowest area?
Proof for the smallest bounding rectangle from @Jaap Scherphuis's answer:
Suppose you start with a filled rectangle, with an area that is yet to be determined. It has 4 sides. We then take away as few cells as possible until the number of sides has reached the value we want.
If you remove a corner cell, the number of sides is increased by $2$.
If you remove any other boundary cell, the number of sides is increased by $4$.
It is impossible to create more than $4$ extra sides by the removal of one cell.
Therefore, if $n$ is a multiple of $4$, the best we could do is to remove $\frac{n-4}{4}$ non-adjacent boundary cells, and if $n$ is even and not a multiple of $4$ the best we could do is remove one corner cell and $\frac{n-6}{4}$ non-adjacent boundary cells.
We want the remaining area to be $n$, so the ideal optimal solution would be to start with a rectangle of area $n+\frac{n-4}{4}$ (if $4|n$) or $n+\frac{n-6}{4}+1$ (otherwise). These two expressions can be combined as $n+\lfloor \frac{n-2}{4}\rfloor$.
This does not always work, unfortunately.
The ideal optimal area sometimes does not make a rectangle from which you can take the required number of cells. In particular, the area could be a prime number, and a rectangle of width $1$ can't have boundary cells removed from it. This happens for example with $n=6$. The ideal would be a rectangle of area 7 with one corner cell removed, but that is impossible. In this case you have to use a rectangle of area $8$, and remove a 2-cell corner.
By enlarging the prime area by $1$ you get an even area, and that will always allow you to make a rectangle from which you can remove the required number of boundary cells.
There are also some non-prime cases that fail, for example $n=136$ has an ideal rectangle area of $169=13\times13$, but there is not enough room to remove 33 non-adjacent boundary cells. Again, by increasing it by 1 you can make a $2\times85$ rectangle from which you can easily take out 34 cells (32 single cells and one adjacent pair). I think that if the ideal area is the product of two primes $pq$ and they satisfy $(p-5)(q-5)>9$, then the construction fails.
So the optimal rectangle area is $n+\lfloor \frac{n-2}{4}\rfloor$, except if that number is prime or the product or two large primes, then you need an area of $n+\lfloor \frac{n-2}{4}\rfloor+1$.
A way to build polygons with asymptotically optimal density for any $n>8$, from @AxiomaticSystem's answer:

Densities: $\frac{4}{5}-\frac{2}{5k}$ for $n=4k-2$, $\frac{4}{5}$ for $n=4k$.
Thank you for your replies!