Using Henry’s notation, the system of recurrences is
$\left\{\begin{align*} &m_{AC}(n+1)=2m_{AC}(n)+m_{CA}(n)+2\\ &m_{CA}(n+1)=m_{CB}(n)+m_{BA}(n)+1\\ &m_{AB}(n+1)=m_{AC}(n)+m_{CB}(n)+1\\ &m_{BA}(n+1)=m_{BC}(n)+m_{CA}(n)+1\\ &m_{BC}(n+1)=m_{BA}(n)+m_{AC}(n)+1\\ &m_{CB}(n+1)=m_{CA}(n)+m_{AB}(n)+1\;. \end{align*}\right.$
Then $\begin{align*} m_{CA}(n+1)&=m_{CB}(n)+m_{BA}(n)+1\\ &=2m_{CA}(n-1)+m_{AB}(n-1)+m_{BC}(n-1)+3\\ &=2m_{CA}(n-1)+2m_{AC}(n-2)+m_{CB}(n-2)+m_{BA}(n-2)+5\\ &=2m_{CA}(n-1)+2m_{AC}(n-2)+m_{CA}(n-1)+4\\ &=3m_{CA}(n-1)+2m_{AC}(n-2)+4\;. \end{align*}$
On the other hand, $m_{CA}(n)=m_{AC}(n+1)-2m_{AC}(n)-2$, so $\begin{align*} m_{CA}(n+1)&=m_{AC}(n+2)-2m_{AC}(n+1)-2\;,\\ m_{CA}(n-1)&=m_{AC}(n)-2m_{AC}(n-1)-2\;, \end{align*}$
and hence $\begin{align*} m_{AC}(n+2)-2m_{AC}(n+1)-2&=m_{CA}(n+1)\\ &=3m_{CA}(n-1)+2m_{AC}(n-2)+4\\ &=3m_{AC}(n)-6m_{AC}(n-1)+2m_{AC}(n-2)-2\;, \end{align*}$
or $m_{AC}(n+2)=2m_{AC}(n+1)+3m_{AC}(n)-6m_{AC}(n-1)+2m_{AC}(n-2)\;.$
Finally, shifting indices gives us $m_{AC}(n)=2m_{AC}(n-1)+3m_{AC}(n-2)-6m_{AC}(n-3)+2m_{AC}(n-4)\;.$
By direct computation the initial values are $m_{AC}(0)=0$, $m_{AC}(1)=2$, $m_{AC}(2)=7$, and $m_{AC}(3)=19$. The sequence continues $47, 113, 267, 629$ and is not in OEIS.
The auxiliary equation is $x^4-2x^3-3x^2+6x-2=0$, which reduces to $(x-1)(x^3-x^2-4x+2)=0\;.$ The cubic factor has three real roots; I didn’t feel like writing down the exact expressions for them, but this cubic equation calculator says that they are approximately $\begin{align*}&2.34292308277717\;,\\ -&1.81360650264833\;,\text{ and}\\ &0.47068341987116\;. \end{align*}$
In the long run the first will dominate, and the sequence will grow by approximately that factor with each increase of one in the number of disks. For the record, $629/267\approx 2.3558$.