Why is $x^n\approx \left(n(x^{1/4096}-1)+1\right)^{4096}$?

For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that $$ \lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n. $$ In fact, we have $$ \ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] = \lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x, $$ where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.

A standard trick is to calculate the natural logarithm first to get the exponent under control:

$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$ Set $u=1/a$. We get $$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$ Use L'Hopital: $$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$ Here we just plugged in $u=0$ to calculate the limit!

So the original limit goes to $x^n$ as desired.

If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case

$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$

If $n$ is also relatively small, then

$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$

Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,

$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$

is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get

$${2.7136203+2.7152613\over2}=2.7144408$$

which is quite close to the true value!