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.