The video shows someone doing large numbers of digits. This is a base-like calculation.
Computers (ie people who can do rapid calculations), can do this with some ease, but it takes practice. I can do several places of criss-cross calculations in various bases, like 120, and 10, but the mind does not hold the digits easily.
I'm pretty sure that some of the super feats done by the Trachenberg system gives the same sorts of results. In any case, the person might have a greater difficulty with a different base, because, like the chinese stone-cage, the process uses features of base 10.
My grandfather could add money columns (£,s,d) as fast as one can run a finger down the column. I suppose, it's not so much the mathematics, or the form of the number, but the kid in the video can just hold hundreds of digits and do lots of simple arithmetic slides.
On Bases
The only 'real' numbers are the digits. Anything else is represented as a path of remainders. So for example, '7' is a real thing, but '73' is a pathway to a number. The same numbers in dozenal are '7' and '61'.
Something like '7!' is a different path to a number. One can for example, learn 7! in ten or dozenal or 120, to get eg 5040, 2E00, and 4200. But these are not 'conversions' or even calculations. They're the outcome of rote-learnt tables. Other numbers do not lend themselves to such fast conversion. A similar-sized 5120, i do not know its dozenal form, but its twelfty form is 4280. I suppose you can write it as 5040 + 80, and this allows one to directly write 2E68 for the dozenal.
In all of the examples above, the numbers represent a series of remainders, but i did not go through the remainders to find the number. I saw 5040 as 7!, and wrote 7! in the various bases.
Some numbers i know only in base 10, some in base 12, some in base 120. The order of a gosset's E8, is 3.43.24.00.00. If you know further that this is 1.72*10!, you could work it out in dozenal, by noting that 10! = 1270000, and that 127 is dec 175, and this divided by 9 gives 19 4/9. From this one gets 1754 00000. The decimal, you could get by multiplying 6912 * 1008 * 100, which i think is 696729600.
Many years ago, i did a project to find the index of primes 2-19 for the primes: this is for a primitive root g, then eg $g^i=2 \pmod p$. This of course, suited the factor-model, because you would see something like, eg 507, and you say, $3.13^2$, and the like. I found it hard going when $p$ got to 56.00 (6720). I could not look at a number like say 3135 and say immediately, that it is 3*5*11*19. (this one i could, but there are others i couldn't that had fairly small divisors.) The factor method is not really something one does.
Some features are learnt through the fingers. A typist might tell that a word is typed wrong because the feeling from the keys is that. I know more that eg 696729600 feels right, rather than looking at the digits. Likewise, i can type in the various short-chords of the polygons, (the chord which makes the third side of a triangle of two edges), usually without thought, eg 1.801937736 or 1.93185165259 for {7} and {12}. I have typed these in quite often. The heptagon was done on a ten-digit calculator, the dodecagon is older, was done on a 12-digit calculator.
Best Answer
Octal multiplication is not that different from regular multiplication. I'm going to work through a different example ($25·47$) in base 10 and then in base 8 so that you can see where the similarities and the differences are.
First, what does $25·47$ really mean? In base 10,
25means $2·10 + 5$ and47means $4·10+7$. So we're really doing: $$(2·10+5)(4·10+7).$$Ordinary algebra tells us that $(a+b)(c+d) = ac + ad + bc + bd$, so we can rewrite this as
$$\begin{array}{lll} \text{hundreds} & \text{tens} & \text{units} \\ \hline \color{darkblue}{2·4}·10^2 & + \color{purple}{2}·10·\color{purple}{7} + \color{purple}{5·4}·10 & + \color{maroon}{5·7} = \\ \color{darkblue}{2·4}·10^2 & + \color{purple}{(2·7+5·4)}·10 & + \color{maroon}{5·7} \end{array} $$
The three terms here represent the hundreds (blue), tens (purple), and units (red) columns. Now we need to do the carrying.
$\color{maroon}{5·7} = 3·10+5$, so we move the three tens from the units column into the next column to the left: $$\begin{array}{lll} \text{hundreds} & \text{tens} & \text{units} \\ \hline \color{darkblue}{2·4}·10^2 & + \color{purple}{(2·7+5·4\color{maroon}{+3})}·10 & + \color{maroon}{5}\end{array}$$
and then $\color{purple}{2·7+5·4 + \color{maroon}{3}} = 3·10+7$, so we move the three tens left from the tens column:
$$\begin{array}{lll} \text{hundreds} & \text{tens} & \text{units} \\ \hline \color{darkblue}{(2·4\color{purple}{+3})}·10^2 & + \color{purple}{7}·10 & + \color{maroon}{5}\end{array}$$
Finally, $ \color{darkblue}{2·4\color{purple}{+3}} = 1·10+1$, so we move the ten left from the hundreds column into a new column:
$$\begin{array}{llll} \text{thousands} & \text{hundreds} & \text{tens} & \text{units} \\ \hline 1 ·10^3 & + \color{darkblue}{1}·10^2 & + \color{purple}{7}·10 & + \color{maroon}{5}\end{array}$$
And in base 10, we abbreviate this to just $$1 \color{darkblue}{1} \color{purple}{7} \color{maroon}{5}$$
which is the answer: $25·47 = 1125$. Now make sure you followed that carefully, because we're going to do it over in base 8.
What does $25·47$ really mean in base 8? In base 8,
25means $2·8 + 5$ and47means $4·8+7$. So we're really doing: $$(2·8+5)(4·8+7).$$We can rewrite this as
$$\begin{array}{lll} \text{64s} & \text{8s} & \text{units} \\ \hline \color{darkblue}{2·4}·8^2 & + \color{purple}{2}·8·\color{purple}{7} + \color{purple}{5·4}·8 & + \color{maroon}{5·7} = \\ \color{darkblue}{2·4}·8^2 & + \color{purple}{(2·7+5·4)}·8 &+ \color{maroon}{5·7}\end{array}$$
The three terms here represent the sixty-fours, eights, and units columns. Now we need to do the carrying.
$\color{maroon}{5·7} = 4·8+3$, so we move the four eights from the units column into the next column to the left:
$$\begin{array}{lll} \text{64s} & \text{8s} & \text{units} \\ \hline \color{darkblue}{2·4}·8^2 &+ \color{purple}{(2·7+5·4\color{maroon}{+4})}·8 &+ \color{maroon}{3}\end{array}$$
and then $\color{purple}{2·7+5·4 + \color{maroon}{4}} = 4·8+6$, so we move the four eights left from the eights column:
$$\begin{array}{lll} \text{64s} & \text{8s} & \text{units} \\ \hline \color{darkblue}{(2·4\color{purple}{+4})}·8^2 & + \color{purple}{6}·8 & + \color{maroon}{3}\end{array}$$
Finally, $ \color{darkblue}{2·4\color{purple}{+4}} = 1·8+4$, so we move the one eight left from the sixty-fours column into a new column:
$$\begin{array}{llll} \text{512s} & \text{64s} & \text{8s} & \text{units} \\ \hline 1 ·8^3 &+ \color{darkblue}{4}·8^2 &+ \color{purple}{6}·8 &+ \color{maroon}{3}\end{array}$$
And in base 8, we abbreviate this to just $$1 \color{darkblue}{4} \color{purple}{6} \color{maroon}{3}$$
which is the answer: $25_8·47_8 = 1463_8$.
Now let's see if we can do the octal multiplication shorthand, without so much toil. We want $$\begin{array}{lll}&2&5\\×&4&7\\\hline \end{array}$$ but with everything in base 8.
We start as usual: $5×7 = 35$… in base 10 we would put down the 5 and carry the 3, but in base 8 we understand it as 4 eights and 3 units, so we put down the 3 and carry the 4:
$$\begin{array}{lll}&2&5^4\\×&4&7\\\hline &&3\end{array}$$
Now $5×4 $ plus the $4$ we carried is $24$, but that's 3 eights and no units, so we put down $30$, not $24$:
$$\begin{array}{lll}&2&5^4\\×&4&7\\\hline 3&0&3\end{array}$$
That $303$ is because $5×47$ is written as $303$ in base 8. Now we continue as usual: $2×7=14$, which is one eight and six units, so we put down the $6$ and carry the $1$:
$$\begin{array}{llll}&&2^1&5^4\\×&&4&7\\\hline &3&0&3\\&&6\end{array}$$
Then $2×4$ plus the $1$ we carried is one eight and one unit, which we put down:
$$\begin{array}{llll}&&2^1&5^4\\×&&4&7\\\hline &3&0&3\\1&1&6\\\hline\end{array}$$
The $116$ is how $2×47$ is written in base 8.
Now we add up the partial products, which is easy; there isn't even any carrying:
$$\begin{array}{llll}&&2^1&5^4\\×&&4&7\\\hline &3&0&3\\1&1&6\\\hline 1&4&6&3\end{array}$$
And 1463 is the answer.