The short answer is in the other comment. It's an exponential problem that with some simplifications can be quite well approximated by a polinomial problem.
About our calculation, with a lot of simplifications ...
You want to know the density of the probability to find an electron in any point is the space near a set of a few atoms.
Instead of using calculating the density as a function of (x,y,z) you can calculate it as a sum of the orbitals that are associated to each atom. Assume 5 orbital per atom like Carbon or Oxygen, and 1 orbital per Hydrogen.
The orbitals may be combined. When you do it by hand, the most common case is called "hybridization", but the computer can choose stranger combinations. It is just an orthonormal base of L^2.
One way to describe the density is to use a matrix of NxN (or 5Nx5N, or whatever is the actual number of orbitals). The diagonal terms d_i^i are the probability that an electron is in the orbital number i. The rest of the matrix is more difficult to define with handwaving, but they are the correct numbers that allow you to change the base of orbitals and get the correct densities in the new base of orbital using the standard matrix formula C M C^-1.
The numbers may be complex number, but with some standard trick you can assume that they are real. They may be positive, zero or negative anyway. Except in the diagonal, because in the diagonal they are probabilities.
This matrix has not all the information. It's better to have a matrix about pairs of electrons. It's a N^2xN^2 = N^4 matrix (or (5N)^2x(5N)^2 or whatever) where the diagonal items D_ij^ij are the probabilities that you can find an electron in the orbital i and then you can find an electron in the orbital j.
You could assume that the matrix for two electrons is just the product of two matrices of one electron D_ij^ij = d_i^i * d_j^j, but it is not true. It's a good approximation and some methods use it and, they are very fast but less accurate.
This N^2xN^2 = N^4 matrix is actually a tensor that is just like a matrix of 4 dimensions. The operations to change the base are more difficult, but don't worry about that. Sometimes we use as a matrix and sometimes as a tensor. Anyway, the diagonal has the probabilities, and the other coefficients have the correct numbers to make the calculation (somewhat) independent of the base of orbitals you have choose.
You can define an equivalent matrix for 3, 4, 5, ... electrons, they are bigger and provide more information. We (most of the time) use just the version for 2 electrons.
Now you have to operate with a matrix of N^2xN^2 = N^4. Just a naïve matrix multiplication is O(N^6). We do more complicated thing that involves lot of multiplications (there are many ways to multiply them because they have many indices to choose). Imagine something like a diagonalization or an inverse. It's very different, but you don't want to do it by hand.
I hope this is enough to justify a O(N^6). The description has a lot of simplifications.
Notes:
I actually forget about the spin, so you have 2x5xN orbitals for Carbon and Oxygen, instead of 5xN. But the (10N)^4 matrix for 2 electron has blocks, so it's better to operate with the blocks instead of using a big matrix with a lot of zeros.
You can get more precision using more orbitals per atom, but as you can imagine, the calculations is slower.
We use a few different methods, so it's more complicated.
At each step, add "some other research group does that, and get good results in some cases".
At the bottom, they're simulating quantum physics on a classical computer. The complexity should be exponential (in time, mainly), yet they've gotten it down to "linear".
