There is no (useful) exact recursion, but there is an approximate one
Since your comment specified that you prefer to use the MAD around the mean, I will proceed for that case. Given an observed sample vector $\mathbf{x}_n$ this form of the the MAD statistic is:
$$\text{MAD}(\mathbf{x}_n) \equiv \frac{1}{n} \sum_{i=1}^n |x_i - \bar{x}_n|.$$
If we define the values $s_{n:i} \equiv |x_i - \bar{x}_{n}|$ we can write $\text{MAD}(\mathbf{x}_n) = \frac{1}{n} \sum_{i=1}^n s_{n:i}$. We also define the corresponding ordered values $0 \leqslant s_{n:(1)} \leqslant \cdots \leqslant s_{n:(n)}$ and the quantity:
$$r_n \equiv \sum_{i=1}^n \mathbb{I} \Big( |x_{n+1} - \bar{x}_{n}| > (n+1) |x_i - \bar{x}_{n}| \Big).$$
Using these statistics we can write the next value of the MAD statistic as:
$$\begin{align}
\text{MAD}(\mathbf{x}_{n+1})
&= \frac{1}{n+1} \sum_{i=1}^{n+1} |x_i - \bar{x}_{n+1}| \\[6pt]
&= \frac{1}{n+1} \sum_{i=1}^{n+1} \Bigg| x_i - \frac{n \bar{x}_{n} + x_{n+1}}{n+1} \Bigg| \\[6pt]
&= \frac{1}{n+1} \sum_{i=1}^{n+1} \Bigg| x_i + \frac{1}{n+1} \cdot \bar{x}_{n} - \frac{1}{n+1} \cdot \bar{x}_{n} - \frac{n \bar{x}_{n} + x_{n+1}}{n+1} \Bigg| \\[6pt]
&= \frac{1}{n+1} \sum_{i=1}^{n+1} \Bigg| (x_i - \bar{x}_{n}) - \frac{1}{n+1} \cdot (x_{n+1} - \bar{x}_n) \Bigg| \\[6pt]
&=\quad \frac{1}{n+1} \sum_{i=1}^{n+1} \Bigg( |x_i - \bar{x}_{n}| - \frac{1}{n+1} \cdot |x_{n+1} - \bar{x}_{n}| \Bigg) \\[6pt]
&\quad + \frac{2}{n+1} \sum_{i=1}^{r_n} \Bigg( \frac{1}{n+1} \cdot |x_{n+1} - \bar{x}_{n}| - s_{n:(i)} \Bigg) \\[6pt]
&=\quad \frac{n}{n+1} \cdot \Bigg[ \text{MAD}(\mathbf{x}_{n}) + \frac{1}{n+1} \cdot |x_{n+1} - \bar{x}_{n}| \Bigg] \\[6pt]
&\quad + \frac{2}{n+1} \sum_{i=1}^{r_n} \Bigg( \frac{1}{n+1} \cdot |x_{n+1} - \bar{x}_{n}| - s_{n:(i)} \Bigg). \\[6pt]
\end{align}$$
This is an exact recursion, but it requires you to keep track of at least as many statistics as the number of sample values, so it is no more parsimonious than keeping track of the whole sample and computing the MAD statistic from scratch.
An approximate recursion: When $n$ is large it is likely that $r_n$ is small, so we may approximate this equation by assuming that the last term is zero. This gives the approximate recursion:
$$\text{MAD}(\mathbf{x}_{n+1})
\approx \frac{n}{n+1} \cdot \Bigg[ \text{MAD}(\mathbf{x}_{n}) + \frac{1}{n+1} \cdot |x_{n+1} - \bar{x}_{n}| \Bigg].$$
This gives you a simple approximating recursive equation that requires you to keep track of the sample mean and the MAD statistic at each iteration. Of course, the recursion is not exact, since it is possible that $r_n > 0$, and in this case there is an additional positive term in the MAD recursion that is ignored in the approximation.
Unfortunately there is no exact recursion that accounts for this that does not require you to keep track of all the values. If you are looking for something that is better than the above approximation, but less onerous than keeping track of all the values in the sample, you could choose some modest upper bound $R$ for $r_n$ (e.g., an upper bound $R=2,3$, etc.) and keep track of the minimum deveiations for the mean up to that order. This would allow you to keep track of a few of the order statistics $s_{n:(1)} \leqslant \cdots \leqslant s_{n:(R)}$ up to the upper bound which would allow you to approximate using the first few terms in the last summation in the recursion. Using this method it is possible to get an approximate recursion that is quite accurate when $n$ is large.
