When you randomly shuffle a deck of cards, what is the probability that it is a unique permutation never before configured?

Your original answer of $\dfrac{3 \times 10^{14}}{52!}$ is not far from being right. That is in fact the expected number of times any ordering of the cards has occurred.

The probability that any particular ordering of the cards has not occurred, given your initial assumptions, is $\left(1-\frac1{52!}\right)^{(3\times10^{14})}$, and the probability that it has occurred is 1 minus this value. But for small values of $n\epsilon$, $(1+\epsilon)^n$ is nearly $1+n\epsilon$. In particular, since $52!\approx 8\times 10^{67}$ and so $\dfrac{3\times10^{14}}{52!}\approx 3.75\times 10^{-54}$ is microscopically small, $1-\left(1-\frac1{52!}\right)^{(3\times10^{14})}$ is very nearly $\frac1{52!}\times (3\times10^{14})$.

There are $52!$ possible orders for a deck of $52$ cards. If a unique order of a deck of $52$ unique cards had been created every second since the big bang, the chances that any two of them were repeated is approximated by $$1-(1-1/52!)^{(10^{17})} = 1.2397999\times10^{-51}\ .$$ To show the size of this number, assume that the same shuffling has taken place every second on one planet orbiting every one of the estimated $10^{24}$ stars in the known universe since the beginning of time. The chances that all of those orders has been unique is still $$99.999999999999999999999999876\%\ .$$ Go shuffle a deck of cards six times and create something truly unique!

Suppose we shuffle a deck and get a permutation p. For each previous shuffling there is a 1-1/52! chance that p doesn't match it. Each previous shuffling is independent, in that regardless of what p and the other permutations are, the chance of p matching the shuffling is 1-1/52! When probabilities are independent we can simply multiple them to find the chance of all the events happening. In this case, the each event is actually a match not happening, so the chance of no matches given n previous shuffles is (1-1/52!)^n. We can then complete the calculations as Michael did.

You're right to question your assumptions and right that the formula you give ($n=$number of shuffles that have been made/$N=$number of possible shuffles) isn't quite right (as others have noted), but unlike the birthday paradox, here the difference works to lower the chances of a match, not raise them. Working with smaller numbers helps with the intuition a bit: suppose that there have been $n$ rolls of a $N=$20-sided die, and you want to know what your chances are of hitting a match to some previous roll. Then a reasonable first approximation for small $n$ is that the probability of a match is $n/20$: this is correct for $n=0$ and $n=1$, and it matches the 'intuition' of having $n$ previous rolls to match against. But bumping the number up to $n=20$ shows the breakdown of the approximation; after 20 rolls, your odds aren't 100% of rolling a number that's already been rolled once, and after 21 rolls they certainly aren't greater than 100%!

The flaw here, of course, is that after $n$ rolls there won't have been $n$ unique numbers rolled; instead, there are likely to already be some duplicates. But it's also clear from thinking about the probability this way that the odds of a match on your next shuffle (or roll) must be less than what the odds would be if all the previous shuffles were unique, and so must be less than the $n/N$ approximation that you use (with $n=3\times 10^{14}$ and $N=52!$).

(There's also a relatively intuitive way of looking at the birthday paradox that explains its 'paradoxical' nature; there you're not trying to match one thing against $n$, but $n$ things against each other - so the correct quantity to use isn't $n$ itself, but instead the number of possible matches, $n(n-1)/2\approx n^2/2$, with each one (heuristically) having a $1$ in $N$ chance of actually matching; this is why you can start expecting a match for a value of $n$ that's proportional to $\sqrt{N}$ rather than for a value of $n$ that's proportional to $N$.)