The answer is no. E.g., let $(A_1,A_2,A_3)$ be any Borel partition of $\mathbb{R}^d$ such that $A_j\ne\emptyset$ for each $j$.
Let $\nu$ be any measure on $\mathbb{R}^d$ such that $\nu(A_j)=1$ for each $j$.
For each $i$ and $j$ in $\{1,2,3\}$, let the density of the measure $\mu_i(\cdot)=P(X\in \cdot,Y=i)$ with respect to $\nu$ equal the constant $\mu_{i,j}$ on the set $A_j$, where
the $3\times3$ matrix $(\mu_{i,j})$ is

$$
\frac1{10}\,\begin{bmatrix}
2 & 1 & 0 \\
1 & 2 & 1 \\
0 & 1 & 2 \\
\end{bmatrix}.
$$
Then ($\mu_1+\mu_2+\mu_3$ is indeed a probability measure,)
$\delta(\mu_1, \mu_2)=\frac1{10}(|2-1|+|1-2|+|0-1|)=\frac3{10}=\delta(\mu_2, \mu_3)$, $\delta(\mu_1, \mu_3)=\frac25$, $f(x)=j$ if $x\in A_j$ (the comparisons between the densities $\rho_i$ do not depend on the measure with respect to which the densities are taken -- as long as such densities exist),
and $p=\frac25$ (in this case, the sum of the off-diagonal entries of the matrix).

Replacing here the previous matrix by
$$
\frac1{60}\,\begin{bmatrix}
21 & 7 & 0 \\
9 & 13 & 0 \\
0 & 7 & 3 \\
\end{bmatrix},
$$
we get the same values of $\delta(\mu_1, \mu_2)$, $\delta(\mu_2, \mu_3)$, and $\delta(\mu_1, \mu_3)$, but $p=\frac{23}{60}\ne\frac25$.

So, $p$ is not a function of $\delta(\mu_1, \mu_2)$, $\delta(\mu_2, \mu_3)$, and $\delta(\mu_1, \mu_3)$.

**Added:** However, one always has the following lower bound on the misclassification probability $p$ in terms of the total variation norms $\delta(\mu_i,\mu_j)=\|\mu_i-\mu_j\|$:
\begin{equation*}
p\ge\frac12-\frac1{2(k-1)}\,\sum_{1\le i<j\le k}\|\mu_i-\mu_j\|. \tag{1}
\end{equation*}
Here we consider the more general setting, where $k\in\{2,3,\dots\}$, $Y$ takes values in the set $[k]:=\{1,\dots,k\}$,
$f(x)=i$ if $x\in\bigcap_{j\in J_i}A_{ij}$, $J_i:=[k]\setminus\{i\}$,
\begin{equation*}
A_{ij}:=\begin{cases}
\{x\colon\rho_i(x)>\rho_j(x)\}&\text{ if }i<j,\\
\{x\colon\rho_i(x)\ge\rho_j(x)\}&\text{ if }i>j,
\end{cases}
\end{equation*}
so that $\bigcup_{i\in[k]}\bigcap_{j\in J_i}A_{ij}=\mathbb R^d$. Here everywhere $i$ and $j$ are in $[k]$.

Thus,
\begin{multline*}
1-p=P(f(X)=Y)=\sum_{i\in[k]}P\Big(Y=i,X\in\bigcap_{j\in J_i}A_{ij}\Big)
=\sum_{i\in[k]}\mu_i\Big(\bigcap_{j\in J_i}A_{ij}\Big) \\
\le\sum_{i\in[k]}\min_{j\in J_i}\mu_i(A_{ij})
\le\sum_{i\in[k]}\frac1{k-1}\sum_{j\in J_i}\mu_i(A_{ij})
=\frac1{k-1}\sum_{1\le i<j\le k}[\mu_i(A_{ij})+\mu_j(A_{ji})] \\
=\frac1{k-1}\sum_{1\le i<j\le k}\int(\rho_i\vee\rho_j)\,d\mu
=\frac1{k-1}\sum_{1\le i<j\le k}\int\frac{\rho_i+\rho_j+|\rho_i-\rho_j|}{2}\,d\mu \\
=\frac1{k-1}\sum_{1\le i<j\le k}\frac{\|\mu_i\|+\|\mu_j\|+\|\mu_i-\mu_j\|}{2}
=\frac12\,+\frac1{2(k-1)}\sum_{1\le i<j\le k}\|\mu_i-\mu_j\|;
\end{multline*}
here we used the fact that $\sum_i\mu_i=\mu$, a probability measure, so that $\sum_i\|\mu_i\|=1$. Thus, we have (1).