Convert 54_eight to base 10.

To convert a number from base $b$ to base 10, we multiply each digit by $b$ raised to the power of its position, starting from 0 for the rightmost digit.

$$54_{\text{eight}} = 5 \times 8^1 + 4 \times 8^0$$ $$54_{\text{eight}} = 5 \times 8 + 4 \times 1$$ $$54_{\text{eight}} = 40 + 4$$ $$54_{\text{eight}} = 44_{\text{ten}}$$

Similarly, convert $134_n$ to base 10.

$$134_n = 1 \times n^2 + 3 \times n^1 + 4 \times n^0$$ $$134_n = n^2 + 3n + 4$$

We are given that $134_n = 54_{\text{eight}}$.
So, we set the base 10 equivalents equal:

$$n^2 + 3n + 4 = 44$$

Subtract 44 from both sides to form a quadratic equation:

$$n^2 + 3n + 4 - 44 = 0$$ $$n^2 + 3n - 40 = 0$$

Factor the quadratic equation. We need two numbers that multiply to -40 and add to 3. These numbers are 8 and -5.

$$(n+8)(n-5) = 0$$

This gives two possible values for $n$:

$$n+8 = 0 \implies n = -8$$ $$n-5 = 0 \implies n = 5$$

Since a number base must be a positive integer and greater than the largest digit in the number (which is 4 in $134_n$), $n=-8$ is not a valid base.
Therefore, $n=5$ is the correct base.

The value of $n$ is $\boxed{5}$.