To calculate $\left(3+1\right)(3^{2}+1)(3^{4}+1)\ldots (3^{64}+1)$, we can cleverly manipulate the expression to simplify it. Let's denote the original expression as $Original$. We start by noticing a pattern that can be exploited by introducing a factor and then subtracting it out:

$$\begin{align*} Original &= \left(3+1\right)(3^{2}+1)(3^{4}+1)\ldots (3^{64}+1) \\ &= \frac{1}{2}\left[\left(3-1\right)\left(3+1\right)(3^{2}+1)(3^{4}+1)\ldots (3^{64}+1)+1\right]-\frac{1}{2} \\ &= \frac{1}{2}\left[(3^{2}-1)(3^{2}+1)(3^{4}+1)\ldots (3^{64}+1)+1\right]-\frac{1}{2} \\ &= \frac{1}{2}\left[(3^{4}-1)(3^{4}+1)\ldots (3^{64}+1)+1\right]-\frac{1}{2} \\ &= \frac{1}{2}\left(3^{128}-1+1\right)-\frac{1}{2} \\ &= \frac{1}{2}\times 3^{128}-\frac{1}{2} \\ &= \frac{3^{128}-1}{2}. \end{align*}$$

Thus, the calculation yields:

$$\boxed{\frac{3^{128}-1}{2}}.$$