Because Boole himself introduced the word "algebra" into the subject.
The term "algebra of logic" appears in Boole's 1854 book on Laws of Thought:
Let us conceive, then, of an Algebra in which the symbols x, y, z, etc. admit indifferently of the values 0 and 1, and of these values alone. The laws, the axioms, and the processes, of such an Algebra will be identical in their whole extent with the laws, the axioms, and the processes of an Algebra of Logic. Difference of interpretation will alone divide them. Upon this principle the method of the following work is established.
Boole strongly emphasized the relation between logic and algebra. References to algebra and its correspondence with logic permeate the book.
Other writers continued to use "algebra of logic" for Boole's system and its later simplification to what is now called Boolean algebra. For example, MacFarlane Principles of the Algebra of Logic (1874), C.S. Pierce "On the Algebra of Logic" (1880), and E. Schroeder Algebra der Logik (1890).
In addition to the analogy that Boole had observed with ordinary algebra, there is an equivalence of Boolean algebras with rings satisfying $x^2=x$ for all $x$, which are equivalent to some algebras (in the modern sense) over the 2-element field.
Best Answer
Consider a sheet of skin stretched into a flat drum head and drummed upon. When the drum head is in vibration, let $f(x,y,t)$ be the height of the drum head at position $(x,y)$ and time $t$. Then $f$ obeys the wave equation: $$\frac{\partial^2}{\partial t^2} f = c^2 \left( \frac{\partial^2}{\partial x^2} f + \frac{\partial^2}{\partial y^2} f \right) \quad (\ast) $$ where $c$ is a physical constant related to things like how tight the skin is stretched and what it is made out of. Such a solution must also obey the physical constraint that there is no motion at the boundary of the drum, where the skin is nailed down.
Every sound can be composed into its overtones. A pure overtone with frequency $\omega$ corresponds to a solution to the wave equation which looks like $f(x,y,t) = g(x,y) \cos(\omega t+b)$ where $$- \frac{\omega^2}{c^2} g =\frac{\partial^2}{\partial x^2} g + \frac{\partial^2}{\partial y^2} g \quad (\ast \ast).$$ Therefore, to understand the sound of a drum, one should figure out for which $\omega$ the PDE $(\ast \ast)$ has solutions which are zero on the boundary of the drum. This is called computing the spectrum of the drum, and a property of the drum which depends only on these $\omega$'s is called a property which one "can hear".
The lowest frequency, which will give the fundamental tone of the drum, will correspond to the lowest nonzero $\omega$ for which $(\ast \ast)$ has solutions. Of course, $(\ast \ast)$ always has the solution that $g$ is a constant and $\omega =0$.
OK, so far that made sense. Now the terminology does something illogical. The name "harmonic" is attached not to the lowest nontrivial frequency, but to the zero frequency. That is to say, $g$ is called "harmonic" if it obeys $$0=\frac{\partial^2}{\partial x^2} g + \frac{\partial^2}{\partial y^2} g \quad (\ast \ast \ast).$$ I don't know the actual history here, but I think of this as a form of mathematical obtuseness. "You musicians want to study the lowest frequency of vibration? Well you can't get lower than zero!"
The actual physical question addressed by $(\ast \ast \ast)$ is "what are the possible stable shapes for a drumhead, if the boundary is not planar? So, if the rim of my drum varies in height, but I tack the drumhead to it anyway, what shape will the drumhead sit at when we're not pounding on it? This is the Dirichlet problem for the Laplace equation; if I give you the values of a harmonic function on the boundary, what does the interior look like?