= Solution
Put $\mathcal C_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to\mathcal C_1$ sends $A$ to $A\times\mathbb N$ with $\alpha_1(a,k)=(a,k+1)$. A <function> $f:A\to UB$ extends uniquely by $(a,k)\mapsto\beta_1^k(f(a))$. Its induced <monad> is $T_0A=A\times\mathbb N$, with unit $a\mapsto(a,0)$ and multiplication $(a,k,l)\mapsto(a,k+l)$. An <algebra for a monad> is determined by the action of $1\in\mathbb N$, so its <Eilenberg-Moore category> is $\mathcal C_1$. Thus this first free–forgetful adjunction is a <monadic adjunction> without needing the given assumption for $n=0$.
The <monadic length> counts successive <Eilenberg-Moore comparison functor> steps until the comparison becomes an equivalence. Fix $0\leq m<n$. For $m\geq1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the <left adjoint> $F_{n,m}$ to the composite forgetful functor $U_{n,m}:\mathcal C_n\to\mathcal C_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of $\mathcal C_m$, unit and multiplication are exactly those of the one-step adjunction:
$$
U_{n,m}F_{n,m}=U_mL_m=T_m
$$
as <monads>, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.
By the assumed one-step <monadic adjunction> property, or the explicit argument above when $m=0$, the <Eilenberg-Moore category> $\mathcal C_m^{T_m}$ is equivalent to $\mathcal C_{m+1}$. Under this equivalence the comparison $\mathcal C_n\to\mathcal C_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using $\alpha_1$, and evaluates the first new value using $\alpha_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole <monadic tower for nested partial unary operations> is
$$
\mathbf{Set}=\mathcal C_0,\quad\mathcal C_1,\quad\ldots,\quad\mathcal C_n,
$$
with the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.
It does not terminate earlier. If $m<n$, take the same underlying two-element <set> in two objects of $\mathcal C_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make $\alpha_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to $\mathcal C_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a <full functor>, and cannot be an equivalence. Therefore
$$
\boxed{\operatorname{monadic\ length}(\mathcal C_n\to\mathbf{Set})=n.}
$$
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