One can draw conformations of n-butane with the carbon-carbon bonds oriented in certain directions and the methyl hydrogens pointing in certain directions that are chiral. However, since rotation about single bonds is typically fast at room temperature, these chiral conformers of n-butane would only be resolvable at extremely low temperatures. Therefore we do not typically consider n-butane to be a chiral molecule.
Take-away lesson: If a chiral conformation can readily interconvert with an achiral conformation, then the molecule cannot be resolved into enantiomers and will be optically inactive.
However, there are molecules where rotation about single bonds is restricted (for example due to steric interference) and the rotational barrier is high enough that the molecules can be resolved into their respective enantiomers at room temperature.
Molecules where the rotational barrier about a single bond is high enough to allow different conformations to be isolated at room temperature are called atropisomers.
The biphenyl pictured below (in a chiral conformation) is an atropisomer due to restricted rotation about the central single bond connecting the two benzene rings. The steric interference that would develop between the 4 ortho groups if the molecule were to approach a planar (achiral) conformation prevents the chiral conformer from equilibrating with the achiral (planar) conformer. Hence this biphenyl has been resolved into its two enantiomers at ambient conditions.
Trans-cyclooctene is also capable of being resolved into its two enantiomers due to restricted rotation about single bonds - it also belongs to the class of atropisomers.
In the figure below trans-cyclooctene conformation (a) is planar, therefore this conformation is achiral. The two conformers pictured in (b) are chiral - these conformations are enantiomers of each other. Trans-cyclooctene will be achiral if the two enantiomers pictured in (b) can readily interconvert through achiral (planar) conformer (a).
Build a model or draw trans-cyclooctene and notice the two hydrogens on the trans double bond; one of them is "inside" the ring in the planar conformation and would cause severe steric problems as the methylene chain tries to flip through the planar conformation to the other face of the double bond as pictured below. This steric interference makes it difficult for the molecule to achieve the planar conformation.
(Image source: John D. Roberts, Marjorie C. Caserio, Basic Principles of Organic Chemistry. Excerpt available at http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/.)
Because of this destabilizing transannular interaction, trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a twisted conformation as pictured above for trans-cyclooctene. In your model of trans-cyclooctene, notice how you have to pull and stretch the methylene chain in order to convert trans-cyclooctene into its mirror image. This rotational movement is highly restricted due to the internal hydrogen. It is this barrier to rotation about singles bonds that allows trans-cyclooctene to be resolved into its two mirror image forms at room temperature.
The barrier to racemize trans-cyclooctene (e.g. pass through the planar conformation (a)) is 36 kcal/mole. As you might imagine, as we make the chain longer it should get easier and easier to loop the methylene chain around the internal (protruding) hydrogen on the double bond. Indeed, the barrier to interconvert the two enantiomers of trans-cyclononene is only 20 kcal/mole. Optically resolved trans-cyclononene must be kept below 0°C to prevent racemization. The barrier to racemization in trans-cyclodecene is even lower at 10 kcal/mole.
Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?
First off, these rings can have chiral conformations, just like butane. I think you are really asking, "Are all higher trans-cycloalkenes optically active as trans-cyclooctene is?". That is, do these larger rings also have difficulty reaching the planar, achiral geometry?
So the answer to your original question is - it depends. It depends on the temperature you specify. At room temperature, trans-cyclooctene is the only trans-cycloalkene where the two enantiomeric conformations are stable enough (e.g. the barrier to reach the planar conformation is high enough) to be isolated. If we lower the temperature to 0°C, then we can isolate the enatiomers of trans-cyclononene. If we go low enough in temperature, we could isolate enantiomers of n-butane.