What is the measure of distance in higher dimensions? In our world we are using kilometers to measure distance. What measurement is used to measure distance in higher dimensions? 
 A: Distance measurements in $n$ dimensional flat space follows the same pattern for $n$ equal 1,2,3, or higher values.
I'm going to assume a straight line, change in position to simplify the math (that is we're measuring what a introductory book would call the "displacement" $s$ rather than distance. But then distance is just an accumulation of many magnitudes of displacement, so the full case follows.
You should already be familiar with the one, two and three-dimensional cases (writing $s_{(n)}$ for the magnitude of displacement in a $n$ dimensional space):
\begin{align*}
s^2_{(1)} &= (\Delta x)^2 \\
s^2_{(2)} &= (\Delta x)^2 + (\Delta y)^2\\
s^2_{(3)} &= (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \,,
\end{align*}
and the extension to higher dimension follows the same pattern (allowing for the choice to label the usual dimensions with the last three letters of the alphabet):
\begin{align*}
s^2_{(4)} &= (\Delta w)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \\
s^2_{(5)} &= (\Delta v)^2 + (\Delta w)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \,.
\end{align*}
For generality, however, we might prefer to label our coordinate directions with a number, so that $x$ is named $x_1$, $y$ is named $x_2$ and so on. That lets us write
$$ s^2_{(n)} = \sum_{i=1}^n (x_i)^2 \,. $$
From this you also get an explanation of why the small, compact dimensions posited in some candidate next generation theories don't effect day-to-day life. Differences of position in those dimensions contribute so little to the total displacement between two points that we can't detect the effects.

This approach can be extended to a Minkowski space (the space of special relativity, by measuring interval in the form
$$ s^2_{(n)+(m)} = \left(\sum_{i=1}^n (ct_i)^2\right) - \left(\sum_{i=1}^m (x_i)^2\right) \,, $$
for a space of $n$ time-like dimensions and $m$ space-like dimensions.
Extension to general relativity requires the introduction of a metric.
A: Whatever unit you're using for distance in 1D is still good in any number of dimensions. Kilometers in manifold of dimension n is fine (assuming non-compactified dimensions).
A: One meter is a unit defined in the "real world" around us – places we can actually visit. Or it is used for the lengths and dimensions of objects we can touch.
It only makes sense to use the same "meter" for other worlds if we can actually get to those worlds. If two worlds are completely separated from each other, it makes no sense to apply the units of one to the other because it would be like mixing apples and oranges.
Imagine a totally separate world with a Gulliver and dwarfs, a world with similar laws of physics as ours but not quite identical – but we can't travel there and back, not even in principle. In that case, there is no answer to the question whether a Gulliver over "there" is greater or smaller than an elephant "here". Magnify the other world or shrink it – all these operations are just ways to visualize the things but they don't change any physics.
So each isolated world has to use its own units which may be and often are in principle inapplicable to other worlds. However, if you discover new dimensions in this world, e.g. by the LHC, it's clear that a meter will still be applicable. It's applicable even for the extra dimensional directions themselves, because of the (local) rotational symmetry between the known dimensions and the extra ones.
On the other hand, in every other world, including those with many dimensions, there exist other units to measure distances. One may use the Compton wavelength of any particle species – if there are massive particles – or the spectral lines if there are any.
In particular, in every gravitating world (e.g. every vacuum in string theory), regardless of the spacetime dimension, one may use the Planck units $\ell_P$ in which the Einstein-Hilbert action has the form
$$ S = \int d^D x  \frac{R}{16\pi \ell_P^{D-2}} $$
Sometimes, the "reduced Planck units" are used. In that case, the $16\pi$ factor is absent:
$$ S = \int d^D x  \frac{R}{\ell_{P,\rm reduced}^{D-2}} $$
Or $16\pi$ may be replaced by $2$.
Again, let me emphasize that $\ell_P$ in every different world/vacuum may have the same symbol but in every different world/vacuum, it denotes a "different thing" and the situations and units may only be considered analogous, not identical.
