mjhjmtu

For 1 qubit, the maximally mixed state is I/2.

So, for two qubits, I assume the maximally mixed state is the maximally mixed state is I/4?
Which is:

$1/4 (|00rangle langle 00| + |01rangle langle 01| + |10rangle langle 10| + |11rangle langle 11|)$

Why is this state more mixed than the following, for instance?

$1/2 (|00rangle langle 00| + |11rangle langle 11|)$

Also, does this generalize to higher dimensions similarly?

For two probability distributions, there is a clear notion how to say which one is more mixed: $vec p$ is more mixed than $vec q$ if it can be obtained from $vec p$ by a mixing process, this is, a stochastic process described by a doubly stochastic matrix (i.e. one which preserved the flat distribution).

Birkhoff's theorem relates this to a concept called majorization, which introduces a partial order on the space of probability distributions.

The same concept generalized to mixed states, allowing us to say which mixed state is more mixed -- for instance, one can establish an order by using the majorization condition on the eigenvalues, and then use a Birkhoff's theorem to prove that one can be converted into the other by a quantum "mixing map" (a unital channel).

This is explained in detail e.g. in http://michaelnielsen.org/papers/majorization_review.pdf, or also in the book of Nielsen and Chuang.

Specifically, this yields that the state with all eigenvalues equal (or equivalently the flat probability distribution) is most mixed.

To relate this to the quantification of mixedness through entropy mentioned in the other answers, the connection comes from the fact that if a state $rho$ is more random than another state $sigma$ in the above sense -- i.e., if $sigma$ can be transformed into $rho$ by mixing, or equivalently the eigenvalues of $sigma$ majorize those of $rho$ -- then the entropy of $rho$ is larger than the one of $sigma$. This property (monotonicity under majorization) is known as Schur-concavity, a property shared e.g. by all Renyi-entropies.

The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ is one bit. For $I/4$ it's two bits of entropy instead.

The entropy of states that only have entries on the diagonal of a density matrix is very easy to compute in general, because you just treat the entries as probabilities and compute the Shannon entropy. The Shannon entropy is maximized when all the probabilities are equal, which is why 0.25 four times beats 0.5 twice with 0 twice.

For a general $d$-dimensional system, the maximally mixed state is the one described by the normalised identity matrix: $$rho=I/d.$$
In the specific case of a two-qubit system, this reduces to the first state you write.

Why is this state more mixed than the following

This is a bit hard to answer, as it depends on what your current understanding/intuition of "more mixed" is. One possible answer is that it is "more mixed" because it represents a state associated with a higher uncertainty, as quantifiable for example via the von Neumann entropy.
You can also have a look at the answers to a similar question on physics.SE for more details.