Write $\|g\|_2^2=\sum_{i=1}^d g_i^2$ and note that the $g_i$ are i.i.d with expectation $1$.
Right tail
Chernoff's bound yields for $s\geq 0$
$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-st} E(e^{s(g_1^2-1)})^d$$
Note that $\displaystyle E(e^{s(g_1^2-1)}) = \frac{e^{-s}}{\sqrt{2\pi}}\int e^{-x^2(\frac 12 -s)} dx$ which is finite only if $s<\frac 12$, and in that case a substitution proves that $\displaystyle E(e^{s(g_1^2-1)}) = \frac{e^{-s}}{\sqrt{1-2s}}$ hence
$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-st} \frac{e^{-sd}}{(1-2s)^{d/2}}$$
This is minimized in $s$ for $s=\frac 12 \frac t{t+d}$ thus
$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-t/2}\left(1+\frac td \right)^{d/2}$$
This is not exactly the bound you're looking for, and it is tighter as $t\to \infty$ provided that $C_2>2$. To get the wanted bound, note that
$$e^{-t/2}\left(1+\frac td \right)^{d/2} = \exp\left(-t^2\left[\frac 1{2t}-\frac{d}{2t^2}\log\left(1+\frac td\right) \right] \right)$$
and $$\frac 1{2t}-\frac{d}{2t^2}\log\left(1+\frac td\right) = \frac 1{2t}\left(1-\frac dt \log\left(1+\frac td\right)\right) $$
Numerically, I observe (and this inequality is referenced on Wikipedia) that $$\forall x\geq 0,\; 1-\frac 1x \log\left(1+ x\right)\geq \frac{1}{2\left(1+\frac 1x \right)}$$
hence
$$\frac 1{2t}\left(1-\frac dt \log\left(1+\frac td\right)\right)\geq \frac{1}{4t\left(1+\frac dt \right)} = \frac{1}{4t+4d}$$
thus finally
$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq \exp\left(-\frac{t^2}{4t+4d} \right)$$
Left tail
Since $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right) = P\left(\sum_{i=1}^d [-g_i^2+E(g_i^2)]\geq t\right)$$ and
$\displaystyle \sum_{i=1}^d [-g_i^2+E(g_i^2)] \leq d$ we have $$t>d \implies P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right) = 0$$ so it suffices to get an upper bound when $t\leq d$.
Chernoff's bound yields $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right)\leq e^{t/2}\left(1-\frac td \right)^{d/2}$$
and numerically $$\forall x\in [0,1),\; 1+\frac 1x \log\left(1- x\right)\leq \frac{1}{2\left(1+\frac 1x \right)}$$ thus
$$t\leq d \implies P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right)\leq \exp\left(-\frac{t^2}{4t+4d} \right)$$
Finally,
$$P\left(|\|g\|_2^2-d|\geq t\right)\leq 2\exp\left(-\frac{t^2}{4t+4d} \right)$$
Best Answer
Yes. Let $X_n\sim N_n(0,I)$ and $Y_n\sim N_n(0,I)$. Since $X_n^TY_n=\sum_{i=1}^nX_{n,i}Y_{n,i}$ and by independence $E[X_{n,i}Y_{n,i}]=0$ and $E[(X_{n,i}Y_{n,i})^2]=1$ for each $i$, applying directly the Central Limit Theorem yields $$ \dfrac{1}{\sqrt{n}}\langle X_n,Y_n\rangle = \dfrac{X_n^TY_n-n\cdot E[X_{n,i}Y_{n,i}]}{\sqrt{n}\sqrt{Var(X_{n,i}Y_{n,i})}}\overset{d}{\longrightarrow}N(0,1),\quad n\to\infty. $$