Let $B_t$ be a 2D Brownian Motion with $B_0 = (1,0)$. Now, express $B_t$ in polars, that is, $B_t = (r(t), \theta(t))$. Let $\tau = \inf\{t > 0 : \theta(t) \geq 2 \pi \}$. What is $\mathbb{P}[\tau \leq 1]$?
In other words, what is the probability that the winding number is at least one in a time less than one?
From the skew-product decomposition of the planar Brownian motion (see for instance Revuz-Yor), it is known that $\theta_t=\beta_{C_t}$ where $C_t=\int_0^t \frac{ds}{\rho_s^2}$ with $\rho_s=\| B_s \|$ and $(\beta_t)_{t \ge 0}$ is a Brownian motion independent from $(\rho_t)_{t \ge 0}$. Consequently we have
$\mathbb{P}(\tau \le 1)=\mathbb{P}(\sup_{t \in [0,1]} \theta_t \ge 2\pi)=\mathbb{P}( \sup_{t \in [0,1]} \beta_{C_t} \ge 2\pi)=\mathbb{P}( \sup_{t \in [0,C_1]} \beta_{t} \ge 2\pi)$ $=\int_0^{\infty}\mathbb{P}( \sup_{t \in [0,\tau]} \beta_{t} \ge 2\pi)p(\tau)d\tau$
where $p(\tau)$ is the density of $C_1$.
The distribution of $\sup_{t \in [0,\tau]} \beta_{t}$ is known and the distribution of $C_1$ has been studied in great details in
MR0576898 Yor, Marc Loi de l'indice du lacet brownien, et distribution de Hartman-Watson. (French) Z. Wahrsch. Verw. Gebiete 53 (1980), no. 1, 71–95.
Putting the pieces together, you will get an integral expression involving the Hartman-Watson distribution.
These two earlier MO questions should shed light on your question:
The former question was answered by Andreas Rüdinger with several relevant references, and this summary: "the root mean square winding number grows logarithmically" with the number of steps.
