What does the @ symbol denote in the beginning of a unix domain socket path in Linux?

The @ probably indicates a socket held in an abstract namespace which doesn't belong to a file in the filesystem.

Quoting from The Linux Programming Interface by Michael Kerrisk:

57.6 The Linux Abstract Socket Namespace

The so-called abstract namespace is a Linux-specific feature that allows us to bind a UNIX domain socket to a name without that name being created in the file system. This provides a few potential advantages:

• We don’t need to worry about possible collisions with existing names in the file system.
• It is not necessary to unlink the socket pathname when we have finished using the socket. The abstract name is automatically removed when the socket is closed.
• We don’t need to create a file-system pathname for the socket. This may be useful in a chroot environment, or if we don’t have write access to a file system.

To create an abstract binding, we specify the first byte of the sun_path field as a null byte (\0). [...]

Displaying a leading null byte to denote such type of a socket may be difficult, so that is maybe the reason for the leading @ sign.

As per man 7 unix

• abstract: an abstract socket address is distinguished by the fact that sun_path[0] is a null byte (\0). All of the remaining bytes in sun_path define the "name" of the socket. (Null bytes in the name have no special significance.) The name has no connection with file system pathnames. The socketâs address in this namespace is given by the rest of the bytes in sun_path. When the address of an abstract socket is returned by getsockname(2), getpeername(2), and accept(2), its length is sizeof(struct sockaddr_un), and sun_path contains the abstract name. The abstract socket namespace is a non-portable Linux extension.

Looks like these are 'abstract' - so no real path is present on filesystem