Does the inner product $\langle \cdot, \cdot \rangle$ induce any other norms other than the 2 norm? In the lecture my professor wrote that the standard inner product on $R^n$ is given by 
$\langle x, y \rangle = x^Ty = \sum\limits_{i=1}^n x_i y_i$ 
which induces a norm $\sqrt{\langle x,x \rangle} = \|x\|_2$
My question is do inner products induce other types of norms...or rather are norms such as the 1-norm or the $\infty$-norm induces by some inner product?

This is a really interesting question, and here is a partial answer.  The $1$ and $\infty$ norms do not come from inner products.  For a norm to have an associated inner product actually gives you a lot of structure.  For example (if the scalars are real for convenience),
$$\left\| x - y \right\|^2 = \langle x - y, x -y \rangle = \langle x, x \rangle - 2 \langle x, y \rangle + \langle y, y \rangle = \left\|x \right\|^2 - 2 \langle x, y \rangle + \left\| y \right\|^2$$
In fact it turns out that there is an identity called the parallelogram law
$$2 \left\|x\right\|^2 + 2\left\|y\right\|^2 = \left\|x + y \right\| + \left\| x - y\right\|$$ 
A norm obeys this identity iff it has an associated inner product.  You can verify that the $1$ and $\infty$ norms do not obey this identity (by finding examples), and therefore cannot have an inner product.  In fact the $p$-norms on $\mathbb{R}^n$ only obey this identity when $p=2$.
Thanks to the comments for some additions.  For a proof of the "iff" claim, see this related question.  If you have a norm which obeys the parallelogram law, you can actually express the inner product directly in terms of the norm by (again real case for convenience)
$$\langle x, y \rangle = \frac{1}{4} \left( \left\| x + y \right\|^2 - \left\| x - y \right\|^2 \right)$$
See here for more information.
