Question: I read that the dimensional formula of distance traveled in $n^{th}$ second is same as that of velocity. Okay, the formula for the distance traveled in $n^{th}$ second is $s\_t= u+\frac{a}{2}(2t-1)$ where $u$ is initial velocity, $a$ is uniform acceleration and $t$ is the time. If we proceed and expand the aforementioned formula we will get $s\_t = u + at-\frac{a}{2}$. Now, the last term i.e. $-\frac{a}{2}$ is not a velocity and isn't the principle of homogeneity violated?

Answer: 
The distance travelled in a time $t$ is:

$$ s = ut + \tfrac{1}{2}at^2 $$

So the distance travelled between $t$ and $t - \Delta t$ is:

$$\begin{align}
 \Delta s &= s(t) - s(t - \Delta t) \\
 &= ut + \tfrac{1}{2}at^2 - u(t - \Delta t) - \tfrac{1}{2}a(t - \Delta t)^2 \\
 &= u\Delta t + \tfrac{1}{2}a(2t\Delta t - \Delta t^2)
\end{align}$$

The equation you cite is obtained by setting $\Delta t = 1$, but remember that you're setting $\Delta t$ equal to **one second** not the dimensionless quantity $1$. So your equation should really be:

$$ \Delta s = u\cdot (1 \space\text{second}) + \tfrac{1}{2}a(2t(1 \space\text{second}) - (1 \space\text{second})^2) $$

or multiplying this out:

$$ \Delta s = u\cdot (1 \space\text{second}) + at(1 \space\text{second}) - \tfrac{1}{2}a(1 \space\text{second})^2 $$

So it is dimensionally consistent.
It is dimensionally correct...

In the derivation, you take ut-u(t-1) which makes it appear as u, but is actually u\*1 second...

Hope this clears your doubt...