How do I find the error of nth iteration in Newton's Raphson's method without knowing the exact root In our calculus class, we were introduced to the numerical approximation of root by Newton Raphson method. The question was to calculate the root of a function up to nth decimal places.

Assuming that the function is nice and our initial value does lead to convergence. Our teacher terminated the algorithm when the two successive iterations had the same first n digits and told us that the the approximation was correct up to the nth digit!

I feel that the termination step is valid if $f(x_n)$ and $f(x_{n+1})$ has different signs but my teacher disagrees. How do I sort this out??

Futhermore how do I find the error in the  nth iteration without knowing the exact root?
Your teacher's example cannot work in general as I present a counter example below.
Nonetheless, I think that your teacher's approach is a reasonable way to explain the intuition behind what happens in a typical case, provided that the proper caveats are given.

I think a more reasonable stopping condition, for programming purposes, is to iterate until the value of $f$ is very small.  If the first derivative is relatively large in a neighborhood of the last iterate, this might be enough to prove that there is definitively a root nearby.  Of course, Christian Blatter has already provided sufficient conditions.

For a counter example, let's suppose that
$$f(x) = x(x-\pi)^2 + 10^{-12}.$$
Then, the Newton's method iteration function is
$$N(x) = x-f(x)/f'(x) =  x-\frac{x (x-\pi)^2+10^{-10}}{3 x^2-4\pi  x+\pi ^2}$$
and if we iterate $N$ 20 times starting from $x_0=3.0$, we get
$$
3., 3.07251, 3.10744, 3.12461, 3.13313, 3.13736, 3.13948, 3.14054, \
3.14106, 3.14133, 3.14146, 3.14153, 3.14156, 3.14158, 3.14158, \
3.14159, 3.14159, 3.14159, 3.14159, 3.14159, 3.14159
$$
Thus, your teacher's method implies there is a root at $x=3.14159$ when, of course, there is no root near here.  There is, however, a root near zero to which the process eventually converges after several thousand iterates.

To place this in a broader context, let's examine the basins of attraction for this polynomial in the complex plane.  There are three complex roots, one just to the left of zero and two at $\pi\pm\varepsilon i$ where $\varepsilon$ is a small positive number.  In the picture below, we shade each complex initial seed depending on which of these roots Newton's method ultimately converges.



Now, it is a theorem in complex dynamics that, whenever two of these basins meet, there are points of the third basin arbitrarily near by.  As a result, there is definitely a number whose decimal expansion starts with $3.14159$ that eventually converges to the root near zero under iteration of Newton's method.
By the quadratic convergence of the algorithm, the corrections decrease very quickly, so that the error is much smaller than the last correction. Usually, you can stop iterations when the latter falls below your tolerance.