From the Wikipedia article on the Discrete Fourier Transform:
The sequence of $N$ complex numbers $x_0, ..., x_{N−1}$ is transformed into an $N$-periodic sequence of complex numbers according to the DFT formula:
$$ X_k=\sum_{n=0}^{N-1} x_n e^{-2\pi ikn/N}.$$
What you have done is taken $N=10$ integers, $n=\{1,2,...,N\}$, and turned them into $10$ time samples $t=\{T_s,2T_s,...,NT_s\}$, or $t_n=nT_s$. (To clarify my notation change, this second set is what you called $n$.)
Then you created the signal $x_n=\text{square}(2\pi ft_n)$ with period $f=(NT_s)^{-1}$and fed it into MATLAB's FFT algorithm, attempting to take the DFT of it (as the FFT is just a fast DFT). However, this is NOT what the FFT expects to see!
Your signal starts with $x_1$ and ends with $x_N$. However, if you see the definition of the DFT I gave above, it expects the signal to start with $x_0$ and end with $x_{N-1}$. Ordinarily, this would not be much of a problem, because if your signal $x$ is actually $N$-periodic, then $x_0=x_N$.
However, as noted in the comments by Jim Clay and Jason R above, the signal you start with is not actually a square wave. As you can see in your screenshot, there are six "1" values and only four "-1" values. The square wave should have an equal number of "1"s and "-1"s. I do not know why the values you put in are not a proper square wave, and I suspect there is some odd detail in how MATLAB implimented the function $\text{square}$. To create a square wave, you should change the line
n = 0.000001:Ts:t; %Generating Samples
to
n = 0:Ts:t-Ts; %Generating Samples
or, even better, to
N=t/Ts;
n=(0:N-1)*Ts;
which makes it clear that you are sampling at integer multiples of your sampling time. The signal $x$ you generate in this way will be equivalent to what Jim Clay has generated in his answer.
As to why your signal has magnitude $\approx 1.2$ instead of $1$, you need to remember how the square wave is defined. From the Wikipedia article on the square wave:
$$x_{\mathrm{square}}(t) =\frac{4}{\pi}\left (\sin(2\pi ft) + {1\over3}\sin(6\pi ft) + {1\over5}\sin(10\pi ft) + \cdots\right ).$$
The first term of this function has frequency $f$ and magnitude $\frac{4}{\pi}\approx 1.27$. If you look at Jim Clay's plot this is exactly the magnitude in bin 2 of the function he has plotted.
Up to bin $N/2+1$, the value that will be plotted in bin $k$ is the coefficient of the term in the square wave with frequency $(k-1)f$. (The $-1$ comes from MATLAB indexing beginning with one rather than zero).
