Is there a quick way to compute the order of this element in $GL(2,\mathbb{Z}/5\mathbb{Z})$?

It really should not be too hard to compute the first couple of powers of this matrix.

But you can also compute its characteristic polynomial and use that $4=-1$ in $\mathbb Z/5$, which makes it

$$T^2+T+1$$

which you might now is the third cyclotomic polynomial, in particular divides $T^3-1$, and so must the minimal polynomial, hence ...

Added: Quite generally, a short consideration of possible Jordan blocks / eigenvalues / characteristic polynomials, together with the fact that $\mathbb F_5$ has a unique quadratic extension $\mathbb F_{5^2}$, shows -- as user Servaes points out in a comment -- that the maximal possible order of any element in $GL_2(\mathbb Z/5)$ is $24$. And indeed this maximal order is attained e.g. by $$\begin{bmatrix} 2&2\\2&1 \end{bmatrix} = 2\cdot Id_2 + \begin{bmatrix} 0&2\\2&4 \end{bmatrix}.$$

To find this and/or see that this has order $24$, note that your element $\xi_3 := \begin{bmatrix} 0&2\\2&4 \end{bmatrix}$ via the first part of this answer basically behaves like a primitive third root of unity. To find a generator of the multiplicative group of $\mathbb F_5(\xi_3) \simeq \mathbb F_{5^2}$, obviously $\mathbb F_5$-multiples of $\xi_3$ will not do. Next I tried $1 + \xi_3$, but that is well-known to be a primitive sixth root of unity. So the next I checked is $2+\xi_3$, where we get

$$(2 + \xi_3)^2 = 4 + 4 \xi_3 + \xi_3^2 = 4+4\xi_3 - \xi_3 - 1 = 3+3\xi_3 = 3\cdot (1+\xi_3).$$

That's a bingo, because $3 \in \mathbb F_5$ is a primitive fourth root of unity, and as just recorded $(1+\xi_3)$ is a primitive sixth one, so their product is a primitive twelfth ($12 =lcm(4,6)$) root of unity, and hence its square root there must be a primitive $24$-th one.

Of course there are many others.