The $X_i$ could be distributed differently, but does not have to be. The idea of an ensemble is just that the distributions are related somehow. A way to think of this is that, $X_i$ could be the distribution of the output from some randomized algorithm given input $i$. Consider, for example, the algorithm that on input $i$ outputs a uniformly random bit-string of length $i$. We can now describe the outputs of this algorithm as the ensemble $X = \{X_i\}_{i \in \mathbb{N}}$, where $X_i$ follows the distribution of uniformly random strings.
Of course in general the algorithm could do all sorts of things. You could also think of $X_i$ as following the distribution of $i$-bit keys for some encryption scheme (generated by a particular key generation algorithm) or whatever you want.
Of course, algorithms do not have to give output of increasing length for increasing input $i$. And the same goes for the corresponding ensemble $X = \{X_i\}_{i \in \mathbb{N}}$, the $X_i$ do not have to follow a distribution of strings of increasing length.
So in your bit-flipping example you could think of $i$ as some input to your bit-flipping algorithm (or protocol), that makes it produce one bit of output. My guess is that your bit-flipping protocol takes some kind of security parameter. Typically the higher the security parameter the closer to uniformly random the output of the bit-flip will be. So you could describe the output of the protocol as the ensemble $X = \{X_i\}_{i \in \mathbb{N}}$, where $X_i$ is the output of the protocol when using security parameter $i$. Now, consider an other ensemble $U = \{U_i\}_{i \in \mathbb{N}}$, where all $U_i$ is a truly random bit. If $U$ and $X$ are computationally indistinguishable according to your definition, this then roughly means that if you use a high enough security parameter then no one will be able to distinguish the output from your bit-flipping algorithm apart from a truly random bit-flip.
Now the the second part of the question. This is kind of a technicality. $1^n$ is a way to write the number $n$ in unary notation. I.e., $1^n$ just means $n$ ones (so, e.g., $1^2 = 11$, $1^3 = 111$ and so on). Why would you do that? Because if you give an algorithm the input $n$ in binary then the input is only of size $log(n)$. Thus, if the algorithm runs in time $poly(n)$ then you have an exponential time algorithm*! However, in cryptography we often want to give an algorithm a security parameter $n$ and then have the algorithm run in time $poly(n)$, but we also want to call such an algorithm poly-time. So by using unary notation, we can cheat a little bit and get what we want. So your bit-flipping protocol would take the security parameter $i$ as $1^i$ and $X_i$ would be the distribution of output on input $1^i$.
*Because in computer science in general when we call something an exponential time algorithm, we mean exponential in the length of the input.
