To solve the given limit problem, we start by analyzing the expression:

$ \lim\limits_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = p $

This limit exhibits the indeterminate form $1^{\infty}$. To handle this form, we use the transformation:

$ p = e^{\lim_{x \to 0} \left( \frac{\tan x}{x} - 1 \right) \frac{1}{x^2}} $

Expanding $\tan x$ using its Taylor series near $x = 0$, we have:

$ \tan x = x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots $

Substituting this expansion into the limit, we get:

$ \frac{\tan x - x}{x^3} = \frac{\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots - x\right)}{x^3} = \frac{\frac{x^3}{3} + \frac{2}{15}x^5 + \ldots}{x^3} $

This simplifies to:

$ \frac{x^3}{3x^3} = \frac{1}{3} $

Thus, the limit becomes:

$ p = e^{\frac{1}{3}} $

Therefore, the expression for $\log_e p$ is:

$ \log_e p = \frac{1}{3} $

Finally, computing $96 \log_e p$:

$ 96 \log_e p = 96 \cdot \frac{1}{3} = 32 $