Applying Itô's formula to solve this problem is kind of overkill. It is well-known that for any Gaussian random variable $X$ with mean $0$ and variance $\sigma^2$ it holds that
$$\mathbb{E}e^{\xi X} = \exp \left( \frac{1}{2} \sigma^2 \xi^2 \right) \qquad \text{for all $\xi \in \mathbb{R}$.}$$
Since $B_T \sim N(0,T)$ this implies
$$\mathbb{E}e^{B_T} = e^{T/2} \qquad \text{and} \qquad \mathbb{E}e^{2 B_T} = e^{2T}.$$
Using that $\text{var}(X) = \mathbb{E}(X^2)-(\mathbb{E}(X))^2$ we get
$$\text{var}(e^{B_T}) = \mathbb{E}(e^{2B_T}) - (\mathbb{E}e^{B_T})^2 = e^{2T}-e^{T}.$$
Regarding your attempt: Using that
$$e^{B_T} = e^{T/2} + e^{T/2} + e^{T/2} \int_0^T e^{B_s-s/2} \, dB_s$$
we find by Itô's isometry
$$\begin{align*} \text{var}(e^{B_T}) &= \mathbb{E} \big[ (e^{B_T} - e^{T/2})^2 \big] \\ &= e^T \mathbb{E} \left| \int_0^T e^{B_s-s/2} \, dB_s \right|^2 \\ &= e^T \mathbb{E} \left( \int_0^T e^{2B_s-s} \, ds \right) \\ &= \int_0^T e^{-s} \mathbb{E}e^{2B_s} \, ds. \end{align*}$$
In order to calculate the remain integral you have to calculate $\mathbb{E}e^{2B_s}$. Since we have already seen in the first part of my answer that we can easily calculate the variance of $e^{B_t}$ if we know $\mathbb{E}e^{2B_t}$, this approach doesn't make any sense, from my point of view.
Some details are missing from the notes. A filtration $(\mathscr{F}_t)_{t\geq0}$ is called admissible if $\sigma(B_s,s\leq t)=:\mathscr{F}_t^B\subseteq \mathscr{F}_t,\,\forall t\geq 0$ and $B_t-B_s\perp \mathscr{F}_s$ for $0\leq s\leq t$. Then, given a Brownian motion $(B_t)_{t \geq 0}$, we have that $B_t$ is a $\mathscr{F}_t$-martingale if $(\mathscr{F}_t)_{t\geq0}$ is admissible. Indeed: $E[B_t-B_s|\mathscr{F}_s]=E[B_t-B_s]=0$ and $E[B_s|\mathscr{F}_s]=B_s$ because $B_s$ is $\mathscr{F}_s$-measurable by admissibility.
Best Answer
You do correctly understand what the notation refers to. It literally means that you have $d$ independent normal random variables as a $d$-tuple. I have never seen the notation $\mathcal N ^{\otimes d}$ before though. It seems strange to me. $X\sim \mathcal N$ just means that random variable $X$ follows the distribution indicated by $\mathcal N$ (normal distribution). The symbol $\otimes$ generally refers to some kind of product. It doesn't make sense to me to do $\mathcal N \otimes \mathcal N$ though. I think I can understand what is intended by it though.
In this particular context, we are working with random variables and stochastic processes. This is essentially part of measure theory. In measure theory there is a concept of product measure. So the particular probability measure on higher dimensional Brownian motion is related to the product measure on $d$ one-dimensional Brownian motions. The product of measures typically uses the $\otimes$ symbol. See here: https://en.wikipedia.org/wiki/Product_measure
Notice that $\times$ is used for the product of the spaces the variable takes values in, e.g. $\mathbb R \times \mathbb R$, but the sigma algebras use the $\otimes$ symbol. The symbol $\mathcal N$ does not represent a set or sigma algebra or a measure (it represents all three simultaneously in fact!), so using $\otimes$ on it feels strange to me. Maybe it is standard to do so though and I just haven't seen it yet.
One thought is that using $\otimes$ might imply the independence of the $d$ individual normal RVs. Higher dimensional Brownian motion isn't simply a bunch of normal random variables as those could have a variety of covariance relationships. Still though, applying it to $\mathcal N$ seems strange to me. As long as it is defined though, it would be fine.
E.g. I can take $(U,X)\sim \mathcal U(0,1) \otimes \text{Exp}(1)$ to mean that $U$ is uniform on $(0,1)$ and $X$ is exponential with rate 1, and they are independent. Again, though, I have personally never seen $\otimes$ used in this way.