m-ary antipodal signaling
Let the received vector $\vec{y}$ be given as $$\vec{y} = \vec{x}+ \vec{w}$$ where $\vec{x}$ can be any of $c_1$ or $c_2$. The ML detection rule for deciding $c_1$ is given as $$ \frac{pdf_Y(y;c_1)}{pdf_Y(y;c_2)} &gt; 1\tag{1}$$  $$pdf_Y(y;c_1) = (2\pi)^{-M/2}(\sigma^{-M})e^{-\frac{\Vert \vec{y} - \vec{c_1} \Vert^2}{2\sigma^2}}\tag{2}$$ similary $pdf_Y(y;c_2)$ is $$pdf_Y(y;c_2) = (2\pi)^{-M/2}(\sigma^{-M})e^{-\frac{\Vert \vec{y} - \vec{c_2} \Vert^2}{2\sigma^2}}\tag{3}$$ using $c_1 = -c_2$ and (2), (3) in (1) taking log of both sides and simplifying we get the detection rule for deciding $c_1$ as $$ \vec{y}^Tc_1 &gt; 0$$.

Now an error will happen when  $c_2$ gets transmitted but $ \vec{y}^Tc_1 &gt; 0$. 

Therefore we have $$ (c_2 + \vec{w})^Tc_1 &gt; 0$$ Now $z= \vec{w}^Tc_1$ is a gaussian random variable with mean 0 and variance $\frac{E_b}{M}M\sigma^2 = E_b\sigma^2$. Therefore we have $z &gt; \Vert c_1\Vert^2$. Now normalizing $z$ with standard deviation of $z$ (to convert it standard normal with standard deviation 1 and mean 0) we have $$\frac{z}{\sqrt(E_b\sigma^2)} &gt; \frac{\Vert c_1\Vert^2}{\sqrt(E_b\sigma^2)}$$ Now $\Vert c_1\Vert^2 = E_b$ therefore we have probability of error as $$Q(\sqrt\frac{E_b}{\sigma^2})$$ since $\sigma^2 = \frac{N_o}{2}$ (per unit bandwidth). Therefore the error probability is $$Q(\sqrt\frac{2E_b}{N_o})$$

The assumptions are that noise in $\vec{w}$ are independent and identically distributed. 

For the simple case of binary antipodal signal we have an energy per bit of $E_b$ and noise in 1 dimension. Now for the extended case we have energy per bit of $E_b/M$. Hence the expression above

Note: If the noise along each dimension is not IID, then we would not have this same expression even though eucledian distance between the signal points remains the same. But you can use the approach I presented here to compute the bit error rate.