hmmm. Thanks for your help everyone. The solution I have come up ( with your help ofcourse :) is as follows:

I have two custom templates:

   my_model_list.html
   my_model_detail.html

Under views.py:

class MyModel(object):
    # ... Access other models
    # ... process / normalise data 
    # ... store data

@staff_member_required
def my_model_list_view(request) #show list of all objects
    #. . . create objects of MyModel . . .
    #. . . call their processing methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_list.html', context, RequestContext(request))
    return HttpResponse(r)

@staff_member_required
def my_model_detail_view(request, row_id) # Shows one row (all values in the object) in detail     
    #. . . create object of MyModel . . .
    #. . . call it's methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_detail.html', context, RequestContext(request))
    return HttpResponse(r)

Under the main django urls.py:

urlpatterns = patterns( 
    '',
    (r'^admin/myapp/mymodel/$', my_model_list_view),
    (r'^admin/myapp/mymodel/(\d+)/$', my_model_detail_view),
    ( r'^admin/', include( admin.site.urls ) )
)

You need to add your admin URL before the URL patterns of the admin itself:

urlpatterns = patterns('',
   url(r'^admin/preferences/$', TemplateView.as_view(template_name='admin/preferences/preferences.html')),
   url(r'^admin/', include('django.contrib.admin.urls')),
)

This way the URL won't be processed by Django's admin.

This is because your other admins are using the default admin.site. You need to totally replace the default admin.site with your own as explained here (you may also want to read this too).

Or you can just do it piggy-style by monkeypatching the default admin.site.get_urls() method:

from django.contrib import admin

_admin_site_get_urls = admin.site.get_urls

def get_urls():        
    from django.conf.urls import url
    urls = _admin_site_get_urls()
    urls += [
            url(r'^my_custom_view/$',
                 admin.site.admin_view(MyCustomView.as_view()))
        ]
    return urls

admin.site.get_urls = get_urls

Legal disclaimer : I won't be held responsible for any kind of any unwanted side-effect of this "solution", including (but not restricted too) your coworkers defenestrating you on the next code review. It's a dirty solution. It's a mess. It stinks. It's evil. You shouldn't do that, really.

It looks like you want to add new admin urls with your own customized views and template: django.contrib.admin.ModelAdmin.get_urls

You can construct those new admins without a new model this way.

Actually it is simpler. Just before urlpatterns in urls.py patch admin urls like that:

def get_admin_urls(urls):
    def get_urls():
        my_urls =  patterns('',
           url(r'^$', YourCustomView,name='home'), 
        )
        return my_urls + urls
    return get_urls

admin.autodiscover()

admin_urls = get_admin_urls(admin.site.get_urls())
admin.site.get_urls = admin_urls

You have 3 options here:

Using third-party packages which provide menu

This is pretty straight forward, there are some good packages out there which support menu for admin, and some way to add your item to the menu.

An example of a 3rd party package would be django-admin-tools (last updated Dec 2021, checked April 2023). The drawback is that it is a bit hard to learn. Another option would be to use django-adminplus (last updated Oct 2019, checked April 2023) but at the time of writing this, it does not support Django 2.2.

Hacking with django admin templates

You can always extend admin templates and override the parts you want, as mentioned in @Sardorbek answer, you can copy some parts from admin template and change them the way you want.

Using custom admin site

If your views are supposed to only action on admin site, then you can extend adminsite (and maybe change the default), beside from adding links to template, you should use this method to define your admin-only views as it's easier to check for permissions this way.

Considering you already extended adminsite, now you can use the previous methods to add your link to template, or even extend get_app_list method, example:

class MyAdminSite(admin.AdminSite):
    def get_app_list(self, request):
        app_list = super().get_app_list(request)
        app_list += [
            {
                "name": "My Custom App",
                "app_label": "my_test_app",
                # "app_url": "/admin/test_view",
                "models": [
                    {
                        "name": "tcptraceroute",
                        "object_name": "tcptraceroute",
                        "admin_url": "/admin/test_view",
                        "view_only": True,
                    }
                ],
            }
        ]
        return app_list

You can also check if current staff user can access to module before you show them the links. It will look like this at then end:

I think there might be a simpler way than writing custom ORMS to get the admin integration you want. I used it in an app that allows managing Webfaction email accounts via their Control Panel API.

Take a look at models.py, admin.py and urls.py here: django-webfaction

To create an entry on the admin index page use a dummy model that has managed=False

Register that model with the admin.

You can then intercept the admin urls and direct them to your own views.

This makes sense if the add/edit/delete actions the admin provides make sense for your app. Otherwise you are better off overriding the admin index or changelist templates to include your own custom actions