Determine the dimensions of velocity (v) and pressure (P).

The dimension of velocity is length divided by time:

$$[v] = [L T^{-1}]$$

Pressure is defined as force per unit area. Force ($F$) has dimensions of mass times acceleration ($F=ma$), so $[F] = [M L T^{-2}]$. Area ($A$) has dimensions of length squared, $[A] = [L^2]$.
Therefore, the dimension of pressure is:

$$[P] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$

For terms to be added or subtracted, they must have the same dimensions. In the expression $\sqrt{P + \frac{1}{n}}$, the terms $P$ and $\frac{1}{n}$ must have the same dimensions.
So, the dimension of $\frac{1}{n}$ must be equal to the dimension of $P$:

$$\left[\frac{1}{n}\right] = [P]$$ $$[n]^{-1} = [M L^{-1} T^{-2}]$$

To find the dimension of $n$, we take the reciprocal:

$$[n] = [M^{-1} L T^2]$$

The dimension of $P + \frac{1}{n}$ is therefore also $[M L^{-1} T^{-2}]$.

Since $[P + \frac{1}{n}] = [M L^{-1} T^{-2}]$, the dimension of the square root is:

$$\left[\sqrt{P + \frac{1}{n}}\right] = \sqrt{[M L^{-1} T^{-2}]} = [M^{1/2} L^{-1/2} T^{-1}]$$

The given equation is $v = \frac{\sqrt{P + \frac{1}{n}}}{x}$.
We can write this in terms of dimensions:

$$[v] = \frac{\left[\sqrt{P + \frac{1}{n}}\right]}{[x]}$$

Rearranging to solve for $[x]$:

$$[x] = \frac{\left[\sqrt{P + \frac{1}{n}}\right]}{[v]}$$

Substitute the dimensions we found:

$$[x] = \frac{[M^{1/2} L^{-1/2} T^{-1}]}{[L T^{-1}]}$$ $$[x] = [M^{1/2} L^{-1/2 - 1} T^{-1 - (-1)}]$$ $$[x] = [M^{1/2} L^{-3/2} T^0]$$ $$[x] = [M^{1/2} L^{-3/2}]$$

The dimensions are:
Dimension of $n$: $\boxed{M^{-1} L T^2}$
Dimension of $x$: $\boxed{M^{1/2} L^{-3/2}}$