A fair coin is tossed $n$ times by two people. What is the probability that they get same number of heads?

The probability John gets $k$ heads is the same as the probability John gets $n-k$ heads since the coin is fair.

So the answer to the original question is equal to the probability that the sum of Tom's and John's heads is $n$.

That is the probability of $n$ heads from $2n$ tosses which is indeed $\frac{1}{2^{2n}}{2n \choose n}$.

As you have noted, the probability is $$ p_n = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{k} = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \frac{1}{4^n} \binom{2n}{n} $$ The middle equality uses symmetry of binomials, and last used Vandermonde's convolution identity.