Fascinating.
The probability of being last is $1/N$ regardless of which guest starts with the gravy.
I had to do the calculation with Markov chains to get the answer. But after getting it I thought about it.
It is a certainty that one of your neighbors will get it before you. You can choose that neighbor by sitting next to the first person, or you can wait patiently until one of your neighbors gets it.
From there, to be the last person the gravy needs to go all the way around to your other neighbor without first getting to you. There is some probability that it does this.
But, actually, everyone has the same situation. One of their neighbors will get it and then there is some probability that it goes all the way around from there to your other neighbor without coming back to you through your first neighbor.
Since everyone is in the same situation, the probability of being the last is the same for everyone and so everyone has a $1/n$ chance to be the last person getting the gravy.
Out of interest, here's how I did the calculation beforehand. First I envisaged a state diagram (Markov) something like this:

Here we start on the middle ring ($1-n$). If we get to the $1$ we move onto the inner ring, which I call $1'-n'$. As you can see, $1'=1$.
Similarly, if we get to $n$, then we move onto the outer ring, labeled $1''-n''$. Again $n''=n$.
This ends well if we get to 0 (gree), and it ends badly if we get to $n'$ or $1''$ (red). In fact, we may as well have $n'=1''$.
So now this leads to the following transition matrix when $n=5$. Here the columns and rows are $(1=1'),2,3,4,(5=5''),2',3',4',2'',3'',4'',0,(1''=n')$.
\begin{equation}
M=\left(
\begin{array}{ccccc|ccc|ccc|cc}
0 &0 &0 &0 &0 &\frac{1}{2} &0 &0 &0 &0 &0 &\frac{1}{2} &0 \\
\frac{1}{2} &0 &\frac{1}{2} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\
0 &\frac{1}{2} &0 &\frac{1}{2} &0 &0 &0 &0 &0 &0 &0 &0 &0 \\
0 &0 &\frac{1}{2} &0 &\frac{1}{2} &0 &0 &0 &0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &\frac{1}{2} &\frac{1}{2} &0 \\
\frac{1}{2} &0 &0 &0 &0 &0 &\frac{1}{2} &0 &0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 &\frac{1}{2} &0 &\frac{1}{2} &0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 &0 &\frac{1}{2} &0 &0 &0 &0 &0 &\frac{1}{2} \\
0 &0 &0 &0 &0 &0 &0 &0 &0 &\frac{1}{2} &0 &0 &\frac{1}{2} \\
0 &0 &0 &0 &0 &0 &0 &0 &\frac{1}{2} &0 &\frac{1}{2} &0 &0 \\
0 &0 &0 &0 &\frac{1}{2} &0 &0 &0 &0 &\frac{1}{2} &0 &0 &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1
\end{array}
\right)
\end{equation}
where the vertical lines are a guide to the eye separating the $1-5$ from the $2'-4'$ from the $2''-4''$ from the two endings.
Then we can either run $M=M^2$ hundred times to calculate $M^\inf\approx M^{2^{100}}$ or do an inversion following the formula here. Either way, we get an absorbing probability of:
\begin{equation}
B=\left(
\begin{array}{cc}
\frac{4}{5} &\frac{1}{5} \\
\frac{4}{5} &\frac{1}{5} \\
\frac{4}{5} &\frac{1}{5} \\
\frac{4}{5} &\frac{1}{5} \\
\frac{4}{5} &\frac{1}{5} \\
\frac{3}{5} &\frac{2}{5} \\
\frac{2}{5} &\frac{3}{5} \\
\frac{1}{5} &\frac{4}{5} \\
\frac{1}{5} &\frac{4}{5} \\
\frac{2}{5} &\frac{3}{5} \\
\frac{3}{5} &\frac{2}{5}
\end{array}
\right)
\end{equation}
The only relevant numbers here are the first 5 rows (corresponding to states $1-5$). The remaining rows are if you start in states $2'-4'$ or $2''-4''$. The left column gives the probability of not being last to get the gravy and the right column gives the probability of being the last to get the gravy.
I ran this for a few different $n$ values and realized that it was a constant. I should have realized there would be something up with the reference to this becoming a classic puzzle.
Very neat puzzle.