Yes. There is a characterization of the multi-spectral radius in terms of the ordinary spectral radius.
Suppose that $A$ is a unital Banach algebra. If $a_{1},\dots,a_{r}\in A$, then define the multi-spectral radius of $a_{1},\dots,a_{r}$ to be
$\limsup_{n\rightarrow\infty}(\sum_{i_{1},\dots,i_{n}\in\{1,\dots,r\}}\|a_{i_{1}}\dots a_{i_{n}}\|)^{1/n}$.
I claim that the multi-spectral radius of $a_{1},\dots,a_{r}$ is the maximum value of a spectral radius $\rho(x_{1}\iota(a_{1})+\dots+x_{n}\iota(a_{r}))$ where $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x_{i}\iota(a_{j})=\iota(a_{j})x_{i}$ for all $i$ and $j$ and where $\|x_{i}\|\leq 1$ for all $i$.
Observe that if $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x_{i}\iota(a_{j})=\iota(a_{j})x_{i}$ for all $i$ and $j$ and where $\|x_{i}\|\leq 1$ for all $i$, then
$$\rho(x_{1}\iota(a_{1})+\dots+x_{r}\iota(a_{r}))$$
$$=\lim_{n\rightarrow\infty}\|(x_{1}\iota(a_{1})+\dots+x_{r}\iota(a_{r}))^{n}\|^{1/n}
=\lim_{n\rightarrow\infty}\|\sum_{i_{1},\dots,i_{n}\in\{1,\dots,r\}}x_{i_{1}}\iota(a_{i_{1}})\dots x_{i_{n}}\iota(a_{i_{n}})\|^{1/n}$$
$$\leq\limsup_{n\rightarrow\infty}[\sum_{i_{1},\dots,i_{r}\in\{1,\dots,n\}}\|x_{i_{1}}\dots x_{i_{n}}\iota(a_{i_{1}})\dots\iota(a_{i_{n}})\|]^{1/n}$$
$$\leq\limsup_{n\rightarrow\infty}[\sum_{i_{1},\dots,i_{r}\in\{1,\dots,n\}}\|x_{i_{1}}\|\dots \|x_{i_{n}}\|\cdot\|\iota(a_{i_{1}})\dots\iota(a_{i_{n}})\|]^{1/n}$$
$$\leq\limsup_{n\rightarrow\infty}[\sum_{i_{1},\dots,i_{r}\in\{1,\dots,n\}}\|\iota(a_{i_{1}})\dots\iota(a_{i_{n}})\|]^{1/n}
\leq\limsup_{n\rightarrow\infty}[\sum_{i_{1},\dots,i_{r}\in\{1,\dots,n\}}\|a_{i_{1}}\dots a_{i_{n}}\|]^{1/n}.$$
From the following proposition, one can choose $x_{1},\dots,x_{r},B,\iota$ such that $\rho(x_{1}\iota(a_{1})+\dots+x_{r}\iota(a_{r}))$ is the multi-spectral radius of $A$.
Proposition: For each unital Banach algebra $A$ and cardinal $\lambda$, there is a Banach algebra $B$ and an isometric embedding $\iota:A\rightarrow B$ along with a subset $X\subseteq B$ where $|X|\geq\lambda$ such that $\|x\|\leq 1$ for each $x\in X$ and such that $$\|(\iota(a_{1})x_{1}+\dots+\iota(a_{r})x_{r})^{n}\|=\sum_{i_{1},\dots,i_{n}\in\{1,\dots,r\}}\|a_{i_{1}}\dots a_{i_{n}}\|$$ whenever
$a_{1},\dots,a_{r}\in A$ and $x_{1},\dots,x_{r}$ are distinct elements in $X$.
Proof: Let $X$ be a set disjoint from $A$ with $|X|\geq\lambda$. Let $B$ be the collection of all sums of the form $\sum_{w\in X^{*}}w\cdot a_{w}$ where $a_{w}\in A$ for all $w\in X^{*}$ and where $\sum_{w\in X^{*}}\|a_{w}\|<\infty$. Define a norm on $B$ where we set $\|\sum_{w\in X^{*}}w\cdot a_{w}\|=\sum_{w\in X^{*}}\|a_{w}\|$.
$B$ is clearly a vector space. Define multiplication by letting
$$(\sum_{u\in X^{*}}u\cdot a_{u})\cdot(\sum_{v\in X^{*}}v\cdot a_{v})
=\sum_{u,v\in X^{*}}uv\cdot a_{u}a_{v}.$$
Observe that multiplication on $B$ satisfies the inequality $\|\mathbf{b}\mathbf{c}\|\leq\|\mathbf{b}\|\cdot\|\mathbf{c}\|$. One can now check that $B$ is a Banach space, but if you don't want to check that $B$ is a Banach space, then the proof still goes through when you replace $B$ with its completion.
The mapping $\iota:A\rightarrow B$ where $\iota(a)=\epsilon\cdot a$ and $\epsilon$ denoted the empty word is a Banach algebra embedding. All the conditions of the theorem are satisfied.
Q.E.D.
From the above proposition, we know that
$$\rho(x_{1}\iota(a_{1})+\dots+x_{r}\iota(a_{r}))=\lim_{n\rightarrow\infty}\|(x_{1}\iota(a_{1})+\dots+x_{r}\iota(a_{r}))^{n}\|^{1/n}$$
$$=\lim_{n\rightarrow\infty}(\sum_{i_{1},\dots,i_{n}\in\{1,\dots,r\}}\|a_{i_{1}}\dots a_{i_{n}}\|)^{1/n},$$
so the limit superior in the definition of the multi-spectral radius is actually just a limit.
Here is a conjecture that I did not have too much time to think about.
Conjecture: Whenever $A_{1},\dots,A_{r}$ are complex matrices, then the multi-spectral radius of $A_{1},\dots,A_{r}$ is the supremum of the spectral radii of matrices of the form $A_{1}\otimes U_{1}+\dots+A_{r}\otimes U_{r}$ where there exists an $n$ where $U_{1},\dots,U_{r}$ are $n\times n$-unitary matrices .